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AP Physics Notes

Table of Contents

AP Physics NotesTable of Contents2/3: Motion & KinematicsOne-Dimensional MotionProjectile MotionRelative Velocity4: The Laws of MotionTypes of ForcesNewton's First Law of MotionNewton's Second Law of MotionNewton's Third Law of MotionFree Body Diagrams & Sums of ForcesFriction5: Work and EnergyWorkEnergySpring/Elastic Potential EnergyPower6: Momentum & CollisionsMomentumImpulseConservation of MomentumTypes of CollisionsGlancing Collisions7: Circular Motion and the Law of GravityAngular DisplacementAngular VelocityAngular AccelerationRelations Between Angular & Linear QuantitiesCentripetal Acceleration/Centripetal ForceUniversal Law of GravitationGravitational Potential EnergyEscape VelocityKepler's Laws8: Rotational Equilibrium & Rotational MechanicsTorqueCenter of Gravity/Center of MassRotational KinematicsRelationship Between Torque and Angular AccelerationMoment of InertiaMoments of Inertia of Extended ObjectsRotational Kinetic EnergyAngular MomentumVector Nature of Angular Quantities9: Solids & FluidsStates of MatterDensityPressureVariation of Pressure with DepthHydraulicsPressure MeasurementsBuoyant Forces & Archimedes' PrincipleFluids in MotionMass/Volume Flow Rate & the Equation of ContinuityBernoulli's Equation10: Thermal PhysicsThe Zeroth Law of ThermodynamicsHeat vs. TemperatureKelvin and Absolute ZeroGas LawsBoltzmann Ideal Gas LawThe Kinetic Theory of GasThe Internal Energy of a Monatomic Ideal Gas11: HeatHeat/Thermal EnergySpecific HeatConservation of Heat Energy: CalorimetryLatent Heat and Phase ChangesTransfer of Heat12: The Laws of ThermodynamicsThe Basics of ThermodynamicsHeat and Internal EnergyWork and HeatThe First Law of ThermodynamicsHeat Engines and The Second Law of ThermodynamicsEfficiencyReversible and Irreversible ProcessesThe Carnot EngineEntropy and Disorder13: Vibrations and WavesHooke's LawGeneral Waves InformationElastic Potential EnergyVelocity as a Function of PositionComparing Simple Harmonic Motion with Uniform Circular MotionPeriod and FrequencyPosition, Velocity, and Acceleration as a Function of TimeMotion of a PendulumDamped OscillationsTypes of WavesFrequency, Amplitude, and WavelengthWave Speed Under TensionSuperposition and InterferenceReflection of Waves15: Electric Forces and Electric FieldsProperties of Electric ChargesInsulators and ConductorsCharging by Conduction (Contact)Charging by InductionCoulomb's LawThe Superposition PrincipleElectric FieldsElectric Field LinesConductors in Electrostatic Equilibrium16: Electrical Energy and CapacitancePotential Difference, Electric Potential, and VoltageElectric Potential EnergyElectric PotentialPotentials and Charged ConductorsThe Electron-VoltEquipotential SurfacesCapacitance and Parallel-Plate CapacitorsCapacitors in ParallelCapacitors in SeriesEnergy Stored in a Charged Capacitor17: Current and ResistanceElectric CurrentCurrent and Drift SpeedResistance and Ohm's LawResistivityTemperature Variation of ResistanceSuperconductorsElectrical Energy and Power18: Direct Current CircuitsEMFResistors in SeriesResistors in ParallelKirchhoff's Rules and Complex DC CircuitsRC CircuitsHousehold CircuitsThe Measurement of Current and VoltageElectricity Review Material19: MagnetismMagnetic FieldsMagnetic Force on a Current-Carrying ConductorTorque on a Current Loop and Electric MotorsMotion of a Charged Particle in a Magnetic FieldMagnetic Field of a Long, Straight Wire and Ampere's LawMagnetic Force Due to Wires With CurrentMagnetic Domains20: Induced Voltages and InductanceInduced EMF and Magnetic FluxFaraday's Law of InductionMotional EMF21: Electromagnetic WavesThe TransformerMaxwell's PredictionsProduction of Electromagnetic Waves by an AntennaProperties of Electromagnetic WavesThe Spectrum of Electromagnetic WavesMagnetism Review Material22: Reflection and Refraction of LightThe Nature of LightReflection of LightRefraction of LightDispersion and PrismsTotal Internal Reflection23: Mirrors and LensesFlat MirrorsImages Formed by Spherical Mirrors (Concave Mirrors)Ray Diagrams for MirrorsAtmospheric RefractionThin LensesCombination of Thin LensesAberrations24: Wave OpticsConditions for InterferenceYoung's Double Slit ExperimentPhase Change due to ReflectionInterference in Thin FilmsNewton's RingsDiffractionSingle-Slit DiffractionSingle Slit vs. Double Slit Interference PatternsDiffraction GratingOptics Review Material26: Special RelativityThe Principle of Galilean RelativityThe Speed of LightEinstein's Principle of RelativityTime DilationSimultaneityLength ContractionRelativistic MomentumRelativistic Addition of VelocitiesRelativistic Energy and the Equivalence of Mass and EnergyEnergy and Relativisic Momentum27: Quantum PhysicsPlanck's HypothesisThe Photoelectric Effect and the Particle Theory of LightX-RaysDiffraction of X-Rays by CrystalsThe Compton EffectThe Dual Nature of Light and MatterThe Davisson-Germer ExperimentThe Uncertainty Principle28: Atomic PhysicsEarly Models of the AtomAtomic SpectraAbsorption SpectraThe Bohr Theory of HydrogenEmission SeriesBohr's Correspondence Principle29: Nuclear PhysicsSome Properties of NucleiCharge and MassNuclear StabilityBinding EnergyRadioactivityThe Three Types of RadiationThe Decay Constant and Half LifeThe Decay ProcessAlpha DecayBeta DecayGamma DecayTop ProblemsChapter 2/3Chapter 4Chapter 5Chapter 6Chapter 7Chapter 8Chapter 9Chapter 10Chapter 11Chapter 12Chapter 13Chapter 15Chapter 16Chapter 17Chapter 18Chapter 19Chapter 20Chapter 21Chapter 22Chapter 23Chapter 24Chapter 26Chapter 27Chapter 28References

2/3: Motion & Kinematics

This course works with only constant acceleration and, therefore, only linear or constant velocity.

One-Dimensional Motion

Velocity is equal to displacement (change in position) over time.

(1)v=Δxt

Acceleration is equal to change in velocity over time.

(2)a=Δvt

When an object moves with constant acceleration, the average acceleration equals the instantaneous acceleration. Therefore, a=vfv0t, which can be rewritten as:

(3)vf=v0+at

This makes sense, as with linear velocity, the final velocity over a given time period will equal the velocity at the starting point, plus the change in velocity (Δv), which can be derived from (2): Δv=at.

With linear velocity, average velocity is equal to displacement over time, but it is also equal to the average of the initial and final velocities.

(4)v=v0+vf2

Going back to (1), we can conclude that v=Δxt=v0+vf2.

This can be rewritten as:

(5)Δx=12(v0+vf)t

This intuitively makes sense, as when calculating for the area under underneath a linear velocity graph segment (which is displacement), this equation gives the formula for that area.

If we go back to (3), we can substitute in vf=v0+at into (5), getting: Δx=12(v0+v0+at)t.

Simplified and standardized, this becomes:

(6)Δx=v0t+12at2

This can also be derived using simple integration, which makes more practical sense:

dvdt=adv=adtdv=adtv=at+CBy assuming that v(0)=v0, we can get the following:v=at+v0Notice how this is the same as equation 3.dxdt=v=at+v0dx=(at+v0)dtdx=(at+v0)dtx=v0t+12at2+CBy assuming that x(0)=x0, we can get the following:x=v0t+12at2+x0We can now subtract x0 from both sides.xx0=v0t+12at2Δx=v0t+12at2

We can also easily visualize it with the following standard gravity function that describes position as a function of time for a projectile:

(7)s(t)=4.9t2+v0t+s0

Notice how this exactly describes (6).

The following equation can be derived to get an equation that does not use time.

a=vfv0tt=vfv0ax=12(v0+vf)tx=12(vf+v0)(vfv0a)
(8)vf2=v02+2ax

This equation feels arbitrary because it is arbitrary. It doesn't make practical sense and it cannot be intuitively visualized. It is purely used to make calculations easier. It can, however, be completely avoided. The following problem will first be solved without the equation, then it will be solved with the equation:


A jet is taking off from the deck of an aircraft carrier. Starting from rest, the jet is catapulted with a constant acceleration of +31ms2 along a straight runway and reaches a velocity of +62ms. Find the displacement of the jet.

Without the use of (8) (notice how this process involves solving for time; (8) skips this step):

a=Δvt31ms2=62ms0mstt=2sdvdt=a=31ms2dv=31ms2dtdv=31ms2dtv(t)=31ms2t0s2s31ms2tdt=62m

With the use of (8):

vf2=v02+2ax(62ms)2=(0ms)2+2(31ms2)xx=62m

Projectile Motion

Using these two equations, the path of a projectile can be modeled neglecting air resistance, given the original direction and magnitude of the projectile. v0 is the original magnitude of the velocity, and θ is the direction.

Usually, for projectile motion, the acceleration is gravity.

(11)a=9.8ms2

Relative Velocity

The velocity of a relative to b is denoted by using the vector vab.

To figure out the velocity relative to something else, all of the intermediate factors have to be eliminated.

(12)vac=vab+vbcvad=vab+vbc+vcd...

Remember that these are vectors.

Another important thing to keep in mind is how to inverse the relative vectors.

(13)vab=vba

4: The Laws of Motion

Types of Forces

These are the forces that always act upon everything in the universe.

All types of forces can be classified as one of the two following types.

Newton's First Law of Motion

Every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

Newton's Second Law of Motion

(14)F=ma

Newton's Third Law of Motion

Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object.

Free Body Diagrams & Sums of Forces

Free body diagrams display all forces acting on a single object. Sums of forces equations can then be written based off of free body diagrams to calculate certain values. The axis of free body diagrams can be titled in any way to ensure the most efficient solving process.

Sums of forces equations equal 0N if the object is not accelerating and equal ma if the object is accelerating, where m is the mass of that one single object alone.

Let xp be all forces in the positive x direction. Let xn be all forces in the negative x direction. Let yp be all forces in the positive y direction. Let yn be all forces in the negative y direction.

(15)Fx=xpxn=maxFy=ypyn=may

Friction

There are two types of friction:

Kinetic friction is always less than static friction for the same surfaces and objects.

Both types of frictional forces are proportional to the normal force on the object.

(16)Fs=μsnFk=μkn

The proportionality constant between friction and normal force is known as the coefficient of friction, and it is denoted by μ. It varies based off of materials in constant, properties of the materials, dryness of the materials, etc.

5: Work and Energy

Energy is the ability to do work.

Energy is never lost, only converted. (This is the law of conservation of energy.)

Work

(17)W=(Fcosθ)x

The unit of work is the joule (J).

If θ=0, because F=ma, the following can be derived:

W=(Fcos0°)xW=FxW=maxvf2=v02+2axax=vf2v022
(18)W=12mvf212mv02

Energy

Kinetic Energy:

Because kinetic energy is the same as the amount of work put into an object's motion if that object was originally at rest (this is due to the law of conservation of energy), we can take the above derived work equation, set W=KE and v0=0, and get the following equation.

(19)KE=12mv2

Another intuitive way to derive kinetic energy is using calculus as follows. It is important to note that Fdx=W1.

F=maFdx=madxFdx=W=KEKE=madxa=dvdtKE=mdvdtdxKE=mdxdtdvdxdt=vKE=mvdvKE=12mv2

Work-Energy Theorem:

(20)W=KEfKE0

Kinetic energy is the work an object is capable of doing in coming to rest.

Potential Energy:

Potential energy is the energy an object has as a result of its position. Because of the law of conservation of energy, we can take the equation W=max and set W=PE. Because potential energy is the energy in an object due to position, a is going to be gravity. We can therefore replace a with g. Finally, for practical purposes, the distance, x, is always going to be the height of the object, so, for easy referencing, we rename x to h to get the following equation.

(21)PE=mgh

There is an intuitive way of thinking about the potential energy in an object. Imagine raising a ball up to 2m. Let's say you put in x amount of work to accomplish this. Where does that work go? Is the energy you put in lost? No (law of conservation of energy), it's not. That energy is now stored in the ball as potential energy. Now imagine letting go of the ball. On its way down, the ball releases the same amount of energy you put into it. While the ball was up at a height of 2m, this energy was stored in the ball as potential energy.

Potential energy is always relative.

Energy is never lost, only converted:

(22)PEi+KEi=PEf+KEf

Spring/Elastic Potential Energy

Energy stored in a compressed spring is called spring or elastic potential energy.

To compress a spring, you must exert a force F to compress it x amount. The ratio of these two amounts is known as the spring constant of a given spring, denoted by k. When pushing on a spring, the spring must push back with the same amount of force. Therefore, we can derive the following. Fs is the force that the spring exerts when compressed. There is a negative, because the direction that the spring pushes or pulls is in the opposite direction as displacement.

(23)Fs=kx

Elastic potential energy is defined as the amount of work a spring is capable of doing. Remember that W=Fx. However, in the case of a spring, the force is dependent on the distance. F is a function of x, so to calculate the total capable work, we can take the following integral (we can omit the negative from F=kx because the direction does not matter when considering the work being done):

(24)PEs=W=Fdx=kxdxPEs=12kx2

A revised version of (22) would therefore be:

(25)PEsi+PEgi+KEi=PEsf+PEgf+KEf

All this is saying is that all types of energy combined initially is the same as all types of energy combined finally due to the law of conservation of energy.

Including the work done by non-conservative forces which add or take away energy2, we can conclude the following (Wnc is the work done by non-conservative forces):

(26)Ei+Wnc=Ef

A spring-related concept that tends to be confusing is the comparison of the work done by gravity and a spring. For example, take the following problem:


A spring of spring constant k=8.75Nm is hung vertically from a rigid support. A mass of 0.500kg is placed on the end of the spring and supported by hand at a point so that the displacement of the spring is 0.250m. The mass is suddenly released and allowed to fall. At the lowest point of the mass, what is the displacement of the spring from its equilibrium position?

In order to solve this problem, we have to relate the work done by gravitywith the work done by the spring. Let’s use two displacements:Let x1 be the original displacement.Let x2 be the final displacement.The mass will come to rest once the work done by the spring is the sameas the work done by gravity.Wg=mgh=mg(x2x1)Ws=12kx2=12k(x22x12)Wg=Wsx1=0.25mmg(x2x1)=12k(x22x12)0.5kg9.8ms2(x20.25m)=128.75Nm(x22(0.25m)2)x2={0.25m,0.87m}0.25m is the original displacement, therefore the solution displacement is 0.87m.

The difficult concept to relate here is: Why does the mass come to a stop when the work done by gravity and the spring are equal?

A common misconception is to assume that the mass would come to rest once the force pulling up and pulling down is the same. With equal forces, however, it is only guaranteed that the acceleration is zero, not the velocity. Therefore, we have to relate something else.

The mass starts at rest and ends at rest, which means that ΔKE=0. The change in kinetic energy, though, is just the work done by the spring plus the work done by gravity. So: ΔKE=Wg+Ws=0. So: Wg=Ws. Because they pull in opposing directions, depending on how you establish your coordinate system, either Wg or Ws will be negative, so: Wg=Ws.

For a more sophisticated explanation, visit this physics stack exchange page.

Power

Power is the rate at which work is applied. Therefore, the derivative of work with respect to time is power.

(27)dWdt=P

Therefore, average power is the average work (or constant work) over change in time.

(28)P=WΔt

The units of power is Js which is known as the watt (W), not to be confused with W (work).

We can expand (28) to get slightly different forms of the equation:

(29)P=FΔxΔt
(30)P=Fv

(30) is the instantaneous power exerted by a force exactly at the time when velocity is v.

6: Momentum & Collisions

Momentum

Momentum is inertia in motion.

(31)p=mv

The units of momentum are kgms or Ns.

Impulse

Taking Newton's second law, F=ma, we can substitute a for ΔvΔt, getting F=mΔvΔt. Multiplying both sides of the equation by Δt, we can get:

(32)FΔt=mΔv

And also:

(33)Δp=mΔv

This pretty much means that mass times the change in velocity is the change in momentum. The change in momentum is called impulse. FΔt is also, therefore, impulse. Using all three of these, we can get the following impulse equation:

Impulse:

(34)FΔt=mΔv=Δp

Impulse can also be defined as a force applied over a certain amount of time. In other words, a force applied to an object for a certain time causes its momentum to change proportionally to the force applied and the time it was applied for.

Conservation of Momentum

Given any collision, the momentum before always equals the momentum after. Momentum is always conserved. Kinetic energy, however, is not always conserved. The conservation of kinetic energy is dependent on the types of collisions...

Types of Collisions

Most real world collisions fall in the category of partially inelastic collisions.

In elastic collisions, it is important to consider that kinetic energy is completely conserved. This fact allows us to solve problems that would otherwise not be solvable. The following equation can be derived:

Momentum is conserved:p1i+p2i=p1f+p2fm1iv1i+m2iv2i=m1fv1f+m2fv2fm1(v1iv1f)=m2(v2iv2f)Kinetic energy is conserved:KE1i+KE2i=KE1f+KE2f12m1iv1i+12m2iv2i=12m1fv1f+12m2fv2fm1iv1i+m2iv2i=m1fv1f+m2fv2fm1(v1i2v1f2)=m2(v2i2v2f2)m1(v1iv1f)(v1i+v1f)=m2(v2iv2f)(v2i+v2f)Divide the kinetic energy equation by the momentum equation:
(35)v1i+v1f=v2i+v2f

All this equation is saying is that the combined velocity of the first object is equal to the combined velocity of the second object.

To envision this, imagine a Newton's Cradle apparatus. Think about the velocities of the 2 spheres on the end before and after the collision. It lines up with (35).

Glancing Collisions

Glancing collisions works with higher dimensions of momentum vectors. This applies when an object hits another one at an angle, and the resulting velocities of both objects are also at different angles.

The general idea is that glancing collisions involve 2-dimensional vectors instead of 1-dimensional vectors.

It is very useful to draw a picture for these types of problems. There are 2 ways to do these types of problems:

  1. Separate the x axis and y axis into their own separate equations using the components of the momentums

  2. Represent all momentum quantities as vectors throughout the solving process

Here is an example problem:


At an intersection, a 1500kg car traveling east at 25ms collides with a 2500kg van traveling north at 20ms. Find the direction and magnitude of the velocity of the wreckage immediately after the collision, assuming that the vehicles undergo a perfectly inelastic collision.

Solving using strategy 1:

x direction:1500kg25ms+2500kg0ms=(1500kg+2500kg)vxvx=9.38msy direction:1500kg0ms+2500kg20ms=(1500kg+2500kg)vyvy=12.5msCombination:v=<vx,vy>=<9.38ms,12.5ms>v=15.6ms@53

Solving using strategy 2:

momentum 1:p1=m1v1p1=1500kg<25ms,0ms>p1=<37500kgms,0kgms>momentum 2:p2=m2v2p2=2500kg<0ms,20ms>p2=<0kgms,50000kgms>Combination:pf=p1+p2pf=<37500kgms,50000kgms>vf=<37500kgms1500kg+2500kg,50000kgms1500kg+2500kg>vf=<9.38ms,12.5ms>v=15.6ms@53

Both of the methods work. Which one to use depends on personal preference.

7: Circular Motion and the Law of Gravity

This portion of the course works with radians. Radians have no units, as they are a ratio between arc length and radius.

When rotating a line segment along the origin of a circle, once the distance covered by the end of the line segment is equal to the radius, the angle rotated is one radian.

Using this definition, we can say that:

(36)360=2π

Remember, radians have no units. Using the above conversion factor, we can convert radians to degrees and vice-versa whenever needed.

Angular Displacement

Angular displacement is defined as the difference between the original angle and final angle (change in angle) in radians. Angular displacement is Δθ.

Angular Velocity

Angular velocity is denoted by ω. It is defined by angular displacement over change in time.

(37)ω=ΔθΔt

The units of angular velocity are s1 because radians don't have units.

Angular Acceleration

Angular acceleration is denoted by α. It is defined by change in angular velocity over change is time.

(38)α=ΔωΔt

The units of angular acceleration are s2.

Relations Between Angular & Linear Quantities

Arc length is usually denoted by s.

Going back to the definition of a radian, we can say the angle in radians is equal to arc length over radius because radians are the ratio of arc length and radius. Therefore:

(39)θ=srs=rθ

This relates linear displacement and angular displacement. Our next goal is to relate linear velocity to angular velocity. To do this we need to derive ΔθΔt because ω=ΔθΔt. Also, note that ΔsΔt=vt. vt is the tangential velocity. Tangential velocity is the instantaneous velocity which, with angular motion, happens to be tangential.3

We can use (39) to arrive here as follows:

θ=srΔθ=ΔsrΔθΔt=ΔsrΔt
(40)ω=vtrvt=rω

The next goal is to relate linear acceleration to angular acceleration. To do this we need to derive ΔωΔt using the same method as in the derivation of (40).

ω=vtrΔω=ΔvtrΔωΔt=ΔvtrΔt
(41)α=atrat=rα

All of these derivation have a direct correlation with each other considering the radius.

Here are all the equations listed:

(42)s=rθvt=rωat=rα

Centripetal Acceleration/Centripetal Force

Contrary to popular belief, centripetal force doesn't push outward, it pushes inward.4 The only reason people think it pushes outward is actually because of inertia. The tendency for an object to not move wants to make it leave angular motion, in order for it to stay in angular motion, there must be a force pushing back or an acceleration toward the center of the circle. This is called centripetal acceleration.

A really important concept to understand is that centripetal acceleration/centripetal force is not some external force that just pulls on an object. It is the resulting force of other forces. For example, when a car is traveling in a circle, the friction in the turned wheels causes there to be a force that pushes inward. This is called centripetal force. There isn't just some external force acting on the car making it go in a circle. If the wheels were straight, the frictional force to provide the centripetal force wouldn't be there, and the car would go straight. Similarly, on a banked curve, the x component of the normal force is what provides the centripetal force.

The derivation of centripetal acceleration is a complex one. However, I will try my best to explain. centripetal acceleration is denoted by ac.

Look at the diagram below for reference on the derivation.

Given a change in time, Δt, the object in question travels from point A to point B.

The magnitude of vi and vf is the same. Let's call the magnitude of both velocities just v.

As seen on the diagram, Δv=vfvi.

The acceleration we're interested in is ΔvΔt. Notice how the numerator is Δv and not Δv. This is because we are only interested in the magnitude, as the centripetal acceleration always points toward the center of the circle.

It is really important to note that the acceleration we're interested in is the instantaneous acceleration, therefore ac=limΔt0ΔvΔt. Also, as Δt0, θ0.

The triangle created by Δv, vf, and vi is similar to the triangle created by both r's and AB as θ0. Because they are similar, in the diagram, we can see θ in 2 places.

As θ0, the arc lengths can be considered straight. Therefore, Δv can be considered equal to the arc length of θ and v. Going back to the arc length formula, Δv=vθ because s=rθ.

We now have Δv, we now need Δt to get acceleration.

θ=ωΔt because ω=θΔt, so Δt=θω.

Since vt=rω, ω=vtr.

In our case, vt=v, so ω=vr. Plugging this into the Δt equation, we get that Δt=θvr=rθv.

We now have Δv and Δt, so we can get acceleration.

ac=ΔvΔtac=vθrθvac=v2θrθ
(43)ac=v2r=rω2

Here's the purely mathematical way to derive it:

ac=limΔt0ΔvΔt=limΔt0vfviθω=limΔt0vfviθvr=limΔt0vfvirθv=limΔt0(vfvi)limΔt0rθv=vθrθv=v2θrθ=v2r

This YouTube video explains how centripetal force works really well.

Centripetal force is is just centripetal acceleration times mass:

(44)Fc=mac=mv2r=mrω2

Universal Law of Gravitation

Every particle in the universe is attracted to every other particle in the universe. The force of attraction between them is called gravity. The force of gravity is directly proportional to the product of the mass of two object and inversely proportional to the square of the distance between their center of masses:

(45)Fg=Gm1m2r2

It might be useful to ask where this equation comes from. Unfortunately, though, this equation doesn't have a derivation.5

G is the proportionality constant, known as the constant of universal gravitation or the universal gravitational constant.

(46)G6.67×1011m3kgs2

Gravitational Potential Energy

With objects near Earth's surface, the equation PEg=mgh works. However, as we move farther away from Earth by significant amounts, gravity starts to lessen, and, in turn, gravitational potential energy changes. We must redefine gravitational potential energy.

The previous equation only described the change in gravitational potential energy. This new definition will get us the total value of gravitational potential energy, which can be defined as: the work needed to bring an object from infinitely far away (where gravity is 0 and GPE is 0) to a distance r away.

Gravitational Potential Energy Diagram

Because gravity naturally brings objects close together, the work required by an external force to bring objects together is negative because it happens naturally.

The force of gravity gets exponentially less as distance increases. The work done by gravity is force multiplied by distance. Remember that W=Fdx. With a constant force, this evaluates to W=Fx. However, with gravity, the force changes with respect to distance. Our distance variable in this case is r, and the formula for force of gravity is (45). Putting everything together, we can get the following equation:

(47)PEg=W=rGm1m2r2dr

The bounds of the integral are from to r because GPE is

the work gravity is capable of doing in bringing an object from infinitely far away (Where gravity is 0 and GPE is 0) to a distance r away.

Solving the integral in (47), we get:

(48)PEg=[Gm1m2r]r=r[Gm1m2r]r=

At r= the force of gravity and gravitational potential energy is 0, so the second half of the equation is 0.

(49)limrGm1m2r=0

Therefore, the formula for total GPE is:

(50)PEg=Gm1m2r

This YouTube video explains the derivation really well.

Escape Velocity

Escape velocity is the lowest possible velocity an object can have in order to escape the gravitational pull of another object.

There are a few things we must note to be able to derive the formula for escape velocity. The total energy of an object is the kinetic energy plus the gravitational potential energy.

As an object moves away, gravity will lessen its velocity, so the kinetic energy will eventually turn into gravitational potential energy. The escape velocity describes the velocity required to completely escape the gravitational pull of an object. In theory, this can only be achieved at an infinite distance.

Therefore, for theoretical purposes, we can assume that the object is capable to moving infinitely far away. At this distance, gravitational potential energy is 0 and all kinetic energy will have been converted to gravitational potential energy, so KE+PEg=0. We can now solve for the escape velocity. Let me be the mass of the object that is escaping and m be the mass of the object being escaped from.

KE+PEg=012meve2+(Gmemr)=0

Solving for ve, we get:

(51)ve=2Gmr

Notice how escape velocity doesn't depend on the escaping object's mass, only the mass of the object it is escaping from.

This YouTube video explains the derivation.

Kepler's Laws

  1. All planets move in elliptical orbits with the Sun at one of the focal points.

  2. A line drawn from the Sun to any planet sweeps out equal area in equal time intervals.

    All this means is that a planet moves faster when it is closer to the Sun but slower when it's farther away.

  3. The square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun.

    All this means is that the orbital period for every planet is dependent on how far away (on average) it is from the Sun.

    This can be derived in equation form.

    Going back to centripetal force and acceleration, the sums of forces equation for an orbiting planet would be (ms is the mass of the sun, mp is the mass of the planet and T is the orbital period):6

    Fc=mpacGmsmpr2=mpv2rv=ΔsΔt=2πrTGmsr2=1r(2πrT)2Gmsr2=4π2rT2

    Rearranging, we get:

    (52)T2=(4π2Gms)r3

    The proportionality constant can be denoted by Ks.

    (53)T2=Ksr3

    Ks can be defined as the following:

    (54)Ks=4π2Gms

8: Rotational Equilibrium & Rotational Mechanics

Torque

Torque is the ability of a force to rotate a body about some axis. It is measured as the product of the force being applied perpendicularly to the axis of rotation. Torque is denoted by τ.

(55)τ=Fr

F is the force and r is the perpendicular radius from the axis of ration to the force.

The units of torque are Nm. Do not put J as it is highly unconventional.

The positive/negative signs of torque are consistent with the signs of radians. Moving up from the positive x axis (counterclockwise) is positive torque, and moving down from the positive x axis (clockwise) is negative torque.

Sums of torque diagrams and equations can be created in the same manner as sums of forces diagrams and equations to solve for unknowns. In rotational equilibrium, the sums of torque is equal to 0Nm. If an object is in rotational equilibrium, it does not matter where you put the axis of rotation.

An intuitive way to think about torque is imagining opening a door. Pushing closer to the hinges requires more force to achieve the same rotation as pushing farther away from the hinges. Therefore, if the force was the same, you would exert less torque when pushing closer to the hinges because the door wouldn't open as much. If you applied the same force to the edge of the door, the door would move a lot more, so the torque exerted would be greater.

Another way to think of torque is the strength/force of rotation.

Center of Gravity/Center of Mass

When referring to objects with a weight, we usually consider the weight to be concentrated at one point in space. This is called the center of gravity or center of mass.

To calculate where the center of mass is, we have to compute a sort-of average on all the points of mass in a system. To start, pick a point that every other point will be relative to.

The center of mass relative to that point in the x direction will then be (Let n be the number of points of mass in the system):

(56)xcg=i=1nmixii=1nmi

Similarly, the center of mass relative to that point in the y direction would be:

(57)ycg=i=1nmiyii=1nmi

Rotational Kinematics

Because of the similarities between linear and rotational quantities, linear kinematic equations hold true for rotational kinematics.

(58)ωf=ω0+αtθ=ω0t+12αt2ωf2=ω02+2αθ

Relationship Between Torque and Angular Acceleration

When the net torque on an object is not 0Nm, it undergoes angular acceleration.

Torque is equal to Fr, and F=ma. If we try to get a derivation for torque by multiplying both sides of the force equation by r, we get:

(59)τ=Fr=mra=mr2α

Moment of Inertia

The moment of inertia is generally thought of as the inertia of rotating objects. For example, a rotating circular object that has more mass located farther away from its origin is harder to start rotating and stop rotating than an object whose mass is closer to the origin. This is measured as the moment of inertia.

Notice the similarities between the two equations:

(60)F=maτ=mr2α

In the linear equation, the mass is what provides the inertia, but in the angular equation, the moment of inertia is provided by mr2. That above equation only works, though, if the rotating object only has mass concentrated at one point which is a distance r away from the origin.

In order to get an equation to accurately model real world objects where there is mass spread out, we have to sum up every point's moment of inertia. This is denoted by I.

(61)I=mr2
(62)τ=Iα=(mr2)α

Even though I and m have similarities, mass is an intrinsic property of an object that does not change, whereas the I of a system depends upon the axis of rotation and the manner in which the mass is distributed.

The full relationship is:

(63)τ=Fr=mra=mr2α=Iα

Moments of Inertia of Extended Objects

To calculate moments of inertia for extended objects, you need integral calculus. For this reason, all the formulas of moments of inertia of extended objects will be given on tests etc.

The following part of this subsection is NOT needed for this course.

If you are curious, though, as to how to actually calculate these moments of inertia, I will walk you through it.

Given a finite number of points with finite masses that make up an object we know that the moment of inertia is equal to the sum of them:

(64)I=mr2

Now, if we take an extended object, each point will be infinitesimally small and would, therefore, have an infinitesimally small mass. The radius, however, will stay the same.

(65)I=r2dm

We cannot directly relate r and dm, but we cannot pull r2 out of the integral because the two variables are dependent on each other.

For one example, let's calculate the moment of inertia of a rod of length L and mass m that rotates about the center.

Because the rod rotates about the center and its length is L, we can get the following limits of integration:

(66)I=L2L2r2dm

To relate mass to linear quantities, we can use the linear mass density of the object, denoted by λ, which is the mass per unit length. Therefore:

(67)λ=mlm=λl

Differentiating both sides and keeping in mind that the linear mass density is a constant, we get:

(68)dm=λdl

dl represents an infinitesimal piece of r, which means that dl=dr. Putting everything together:

(69)I=L2L2r2λdr

Evaluating the integral, we obtain:

(70)λr33|L2L2=λ(L2)33λ(L2)33

Remember that λ=ml, and also in this case l=L.

(71)mL(L2)33mL(L2)33=23mLL38=112mL2

This was the derivation of the moment of inertia of a rod of length L and mass m that rotates about the center.

(72)I=112mL2

Other objects' moment of inertia can be derived in very similar ways. This was just one example.

Rotational Kinetic Energy

Remember the original kinetic energy formula, KE=12mv2. As we've seen before with moment of inertia, for rotating objects, we must consider the sum of all points with mass. Let the rotational kinetic energy of an object be the sum of the kinetic energy of all points with mass m and tangential velocity v.

(73)KEr=12mv2

Converting tangential velocity to angular velocity using v=rω, we get:

(74)KEr=12m(rω)2=12mr2ω2

The sum of angular velocities is just ω and can even be considered a constant, so we can take it out of the sum.

(75)KEr=12(mr2)ω2

Remember the equation for moment of inertia, I=mr2. Substituting this in, we get:

(76)KEr=12Iω2

Comparing this to the original kinetic energy equation, this should not be surprising considering the similarities we've seen in the past. Going from translational to rotational, I usually takes the place of m and ω usually takes the place of v.

Angular Momentum

Keep in mind the similarities between m and I, and v and ω. Angular momentum is denoted by L.

Because translational momentum is p=mv, angular momentum is therefore (considering the similarities):

(77)L=Iω

Unlike translational moment, whose units are kgms, the units of angular momentum are kgm2s.

Angular momentum also means there is rotational impulse. It works very similarly to translational impulse.

(78)τ=Iα=I(ΔωΔt)=IΔωΔt=ΔLΔt

Replacing α with ΔωΔt, we obtain IΔω in the numerator, which just represents change in L because:

(79)L=IωΔL=IΔω

Putting everything together as seen in (78), we obtain:

(80)τΔt=IΔω=ΔL

This is really similar to translational impulse. All this is saying is that, given an applied torque over a period of time, the angular momentum will change accordingly.

Conservation of momentum laws apply. (YouTube video example)

For single point particles with a known tangential velocity and radius, the angular momentum can be found using the following formula:

(81)L=mvr

This is because L=Iω, and, for single point particles, I=mr2. Converting ω=vr and putting everything together: L=Iω=mr2vr=mvr.

Vector Nature of Angular Quantities

Until now, we've considered only the magnitudes of angular quantities like ω, α, and L. For generalization purposes that don't depend on point of view, they can be considered vectors. The direction of the vector points along the axis of rotation.7

The direction of this vector can be found using the right hand rule. For this, you must use your right hand. Stick your thumb out and curl the rest of your finger. The direction that the rest of your fingers point should match the direction the object is rotating. The direction your thumb points is the direction of the vector. This works for angular velocity, angular acceleration, and angular momentum

9: Solids & Fluids

States of Matter

Density

Density is defined as mass per volume. The more mass something has given the same volume, the more dense it is. Density is denoted by ρ.

(82)ρ=mV

The units of density are kgm3.

The density of water is known to be 1gcm3. This is not in SI unit form. Converting this to SI unit form, we obtain the density of water to be 1000kgm3.

The specific gravity of a substance is the ratio of its density to the density of water at 4C (This is just exactly where the density of water is exactly 1000kgm3). For example, if the specific gravity of a substance is 3.7, its density is 3700kgm3.

Pressure

Pressure is the force that's exerted over an area. The more concentrated the force (so the smaller the area it's spread out over), the greater the pressure. Pressure is denoted by P.

(83)P=FA

The units of pressure are Nm2 which is equivalent to Pa (pascals).

Variation of Pressure with Depth

Within a fluid, all points at the same depth must have the same pressure as long as the fluid is in static equilibrium (it's not moving).

Look at the following figure:

Variation of Pressure with Depth Diagram

The force pushing on top is F1 and the force pushing on the bottom is F2. The cylindrical cutout in the center is just a representation of some portion of the fluid within the bigger container.

Creating a sums of forces equation for the cutout portion of fluid using the free body diagram, we get:

(84)Fy=F2mgF1=0N

It equals 0N because the fluid is in static equilibrium. Going back to the pressure equation, we know that F=PA. Therefore, we can write:

(85)P2AmgP1A=0N

Going back again to the density equation, we know that m=ρV. Volume, however, is just area times height, so m=ρAh. Substituting this in, we get:

(86)P2AρAhgP1A=0N

Dividing each term by A and isolating P2, we finally get:

(87)P2=P1+ρgh

Look back at the diagram. This equation states that the pressure between the top and the bottom of a fluid, separated by a height h is the pressure at the top plus the density times gh. If the top of the fluid is exposed to the atmosphere, then the pressure at the top is 1atm. P2 is always greater than P1.

The shape of the fluid container does not matter for this principle.

Pascal's Principle states that pressure applied to an enclosed fluid transmitted undiminished to every point of the fluid and to the walls of the containing vessel. All this states is that pressure always pushes evenly outward in all directions.

Hydraulics

In hydraulics, a U-shaped tube (called a u-tube) is used to be able to apply larger forces when a smaller force is initially put in.

Remember, the pressure at any same depth of a liquid is always the same, no matter the shape of the container.

Pressure Measurements

Look at the following diagram.

Pressure Measurements Diagram

Our goal is the measure the pressure of the supplied gas.

PA is the pressure of the gas and P0 is the atmospheric pressure. We know that PA>P0 because of how the liquid is displaced.

Because the pressure at any same depth of a liquid is always the same, we know that the pressure at point A is the same as the pressure at point C. Therefore, the pressure at point C is PA.

Using the variation of pressure with depth equation, we know that the pressure at the bottom of the displaced liquid is P0+ρgh.

Therefore:

(88)PA=P0+ρgh

Buoyant Forces & Archimedes' Principle

The force that makes objects float is called a buoyant force. This force exists because pressure is larger at greater depths.

Given an object in a fluid, the pressure pushing downward on top will be less than the pressure pushing upward on the bottom. Therefore, the upward force exceeds the downward force and the object floats.

(89)FB=P2AP1A=(P2P1)A

Because P2P1=ρgh, we get:

(90)FB=ρghAV=AhFB=ρgVm=ρVFB=mg

This means that the buoyant force is equal to the weight of the fluid that is displaced (the space that the object's volume takes up where there would be fluid).

Archimedes' Principle states that:

(91)FB=Wfluid=ρfluidVfluidg

Fluids in Motion

Below, only the highlighted portions are what we work with in this course. The rest are useful to know for vocabulary.

Types of Fluid Flow

Compressibility of Fluids

Viscosity of Fluids

Rotatability of Fluids

An incompressible, non-viscous fluid is called an ideal fluid.

When fluids flow, streamlines are often used to represent the motion of the fluid.8

Steady flow is often called streamline flow.

Mass/Volume Flow Rate & the Equation of Continuity

The equation of continuity states that the mass of fluid entering one end of a vessel must equal the mass exiting the other end of the vessel.9

Mass Flow Rate must remain constant.

Imagine a pipe with fluid flowing through from point 1 to point 2:

Given a change in time, Δt:

Knowing the cross sectional areas:

Because ρ=mV, m=ρV:

Dividing both sides of the equations by time, we get ΔmΔt which is flow rate.

Because the mass flow rate must remain constant, we arrive at:

(92)ΔmΔt=ρ1A1v1=ρ2A2v2

However, in this course we only work with incompressible fluids, meaning the density always remains constant.

Therefore, (92) can be simplified into ΔVΔt, the volume flow rate. Volume flow rate is denoted by Q.

(93)Q=ΔVΔt=A1v1=A2v2

All this states is that given an incompressible fluid, the volume of fluid entering the vessel in a certain time must be the same as the volume of fluid leaving the vessel.

Bernoulli's Equation

Given an incompressible, non-viscous fluid, whenever a fluid flowing in a horizontal pipe encounters a region of reduced cross sectional area (thinner pipe), the pressure of the fluid drops.

Bernoulli's Principle: Where the velocity of a fluid is high, the pressure is low, and where the velocity of a fluid is low, the pressure is high.10

If the pipe has an elevation difference from low to high, given the width of the pipe does not change, the the pressure at the lower elevation must be greater than the pressure at the higher elevation to be able to keep the fluid moving upward.

Look at the following diagram:

Bernoulli's Equation Diagram

This diagram shows a pipe that has an elevation change and a width change.

Remember that F=PA. Because the pressure at the bottom is greater, the force required to get the fluid from the lower section to the higher section would be F1F2. This is just P1A1P2A2.

Also remember that W=FΔx. Putting this together, we obtain that the total work done to make the fluid move from the lower section to the higher section is W=(P1A1P2A2)Δx. For now, let's say A1=A2=A. I will come back to this later. If this is the case, though, we can write that: W=(P1P2)AΔx.

Now, going back to the figure, notice how, because of the equation of continuity, A1dx1=A2dx2. This means that at both instances, the volume is the same. This is why I was able to say that A1=A2, because going back to the equation we derived, no matter the area, AΔx will always result in a constant volume. Therefore, we can write:

(94)W=(P1P2)V

This can be considered the non-conservative work done on the fluid.

Remember that Ei+Wnc=Ef. Rearranging, we get that Wnc is the difference in energies. So:

(95)Wnc=ΔKE+ΔPE=12mv2212mv12+mgh2mgh1

Because Wnc=(P1P2)V, we get:

(96)(P1P2)V=12mv2212mv12+mgh2mgh1

Dividing each side by V, we get:

(97)P1P2=12ρv2212ρv12+ρgh2ρgh1

Putting all the 1s on one side and all the 2s on the other, we finally arrive at Bernoulli's Equation:

(98)P1+12ρv12+ρgh1=P2+12ρv22+ρgh2

This YouTube video dives more in depth into how this equation works and applications of the equation.

10: Thermal Physics

The Zeroth Law of Thermodynamics

Some materials are better than transmitting heat than others. Get a metal tray and a cardboard box out of a freezer. The metals feel colder even though they are at the same temperature. This is because the metal is better at transmitting heat and sucks the heat from your hand faster than the cardboard does in order to achieve thermal equilibrium.11

The Zeroth Law of Thermodynamics states that if bodies A and B are separately in thermal equilibrium with a third object, C, then A and B will be in thermal equilibrium with each other if placed in thermal contact.1213

Heat vs. Temperature

Temperature is a measure of the average kinetic energy of the particles in an object. Heat is the total kinetic energy of the particles in an object.

Heat is a quantity that measures the total energy that is present in an object in the form of particles moving (it is dependent on how much of an object there is) while temperature is local (It doesn't depend on how much of the object you have).

For example, let's say you have a teacup that's filled with boiling water and a bucket that's filled with lukewarm water. The temperature of the teacup is greater because the particles move faster, so their average kinetic energy is greater. However, the heat of the bucket could be greater because, even though the particles move slower, there are a lot more of them, so the total kinetic energy is more.

Kelvin and Absolute Zero

Normal everyday thermometers are limited by the temperatures that they are capable of measuring. Most of them consist of a tube with a liquid that expands or contracts to display the corresponding temperature. A better type of thermometer uses gas and pressure to determine temperature. These are called constant volume gas thermometers.

The lower the temperature of something becomes, the slower the particles in the object move. At one point, however, they stop moving. At that point, they cannot get any colder. This is a temperature known as absolute zero. Absolute zero is 273.15C. Absolute zero can be calculated by looking at the pressure vs. temperature graph of an object, and approximating where the pressure would be negative:

Absolute Zero Pressure vs. Temperature Graph

In order to better be able to make calculations, in science, we often use the Kelvin scale. One Kelvin (K) has the same value as one degree Celsius, except kelvin starts at absolute zero. Therefore:

(99)K=C+273.15

In the imperial unit system, Fahrenheight is used.

(100)F=C×1.8+32

Gas Laws

Gases that behave ideally make calculation a lot easier. Ideal conditions are low pressure and high temperature. Luckily, this is true for most places on Earth, so the equations hold up well.

Avogadro's Number - The number of items in a mole, denoted by NA.

(101)NA=6.02×1023mol1

Boyle's Law - Pressure is inversely proportional to volume. Boyle made the observation that as you decrease the volume of a gas, the pressure increases: P1V. Because they are proportional, there must exist a proportionality constant, which can be written as P=k1V. Rearranging, we obtain k=PV. Because the proportionality constant does not change, when a gas undergoes a change, the product of pressure and volume initially with the equal to the product of the pressure and volume after:

(102)P1V1=P2V2

Charles's Law - Temperature is directly proportional to volume. Charles made the observation that, as the temperature of a gas increases, so does the volume. VT, so V=kT and k=VT.

(103)V1T1=V2T2

Avogadro's Law - Moles are directly proportional to volume. Avogadro made the observation that, the more moles you have of something, the more space it takes up. Vn, so V=kn and k=Vn.

(104)V1n1=V2n2

Combined Gas Law or Ideal Gas Law - a combination of all three laws, which includes moles.

The three proportionalities are:

(105)V1PVTVn

Combining the three proportionalities into one, we get:

(106)VnTP

To make them equal, we need a proportionality constant. We will denote this using R.

(107)V=RnTP

Putting this into standard form, we get that the Ideal Gas Law is:

(108)PV=nRT

or

(109)P1V1n1T1=P2V2n2T2

R is known as the universal gas constant.

(110)R=8.31JmolK=0.0821LatmmolK

Boltzmann Ideal Gas Law

Remember that, according to Avogadro's number, the number of moles if equal to the number of items divided by Avogadro's number.

(111)n=NNA

Substituting this in for n in the ideal gas law, we get:

(112)PV=NNART

Both R and NA are constants, so they can be combined into one constant, kB=RNA. This is known as the Boltzmann constant.

(113)PV=NkBTkB=1.38×1023JK

The Kinetic Theory of Gas

Because gases contains millions of molecules, it helps us to look at their average speeds.

For the kinetic theory of gas to work, we make the following assumptions:

  1. The number of molecules is large, and the average separation between them is large compared with their dimensions. This means that the molecules occupy a negligible volume in the container.

  2. The molecules obey Newtons Laws of Motion, but as a whole, they move randomly.

  3. The molecules undergo elastic collisions with each other and with the walls of the container. Thus, in the collisions both kinetic energy and momentum are conserved.

  4. The forces between molecules are negligible except during a collision.

  5. The gas under consideration is a pure substance; that is, all molecules are identical.

Pressure of a gas is caused by its molecules hitting the sides of the container. Given a cubical container whose side length is L and an ideal gas composed of N particles, let's consider a single particle of the gas.

While approaching the wall of the container, the particle has a a velocity +v and a momentem +mv. After colliding with the wall and undergoing an elastic collision, its velocity is v and its momentum is mv.

Let t be the time it for the particle to go from one wall to the other, and back. Over this period of time, it covers a distance of 2L. Because v=xt and t=xv, we know that:

(114)t=2Lv

Also, remember that FΔt=Δp, meaning that:

(115)F2Lv=(mv)(mv)

Solving for F, we get:

(116)F=mv2L

Because this is the force exerted by the gas, we know that the container exerts an equal but opposite force. Therefore, the force exerted by the container on the gas is:

(117)F=mv2L

Remember that there are N particles, and, because we are in three dimensions, approximately 13 of the particles strike the given wall at a given time. Therefore, the total force is:

(118)F=N3mv2L

Note that v2 was replaced with v2 because, given N particles, we consider the average velocity.

v2 is called the root-mean-square speed: vrms. Substituting this in, we get:

(119)F=N3mvrms2L

Remember that P=FA. Therefore:

(120)P=FL2=1L2N3mvrms2L

V=L3, so:

(121)PV=32N(12mvrms2)

or

(122)PV=32N(KE)

Also, remember that PV=NkBT, so:

(123)NkBT=32N(KE)

Isolating KE, we get:

(124)KE=32kBT=12mvrms2

The Internal Energy of a Monatomic Ideal Gas

The internal energy U of a gas is the total translational kinetic energy of the N particles in the gas.

(125)U=N(KE)

This can be rewritten as:

(126)U=N(32kBT)

Usually, U is expressed in terms of the number of moles n, rather than the number of atoms N. Remember the following two equations:

(127)kB=RNAn=NNA

Substituting these into the previous equation and simplifying, we obtain:

(128)U=32nRT

11: Heat

Heat/Thermal Energy

There are two types of heat, mechanical energy (KE and PE) and heat/thermal energy. Heat energy can be transformed into mechanical energy.

Heat/thermal energy is energy that is transferred between a system and its environment because of a temperature difference between them.

One of the most widely used units for heat is cal (calories). 1cal is the amount of heat energy required to raise the temperature of 1g of water by 1C.

Calories in food are measure by Calories (specifically a capital C).

(129)1Cal=1kcal=1000cal

In this course, however, because heat is a form of energy, we will use J.

Joule's Experiment

Joule set up an apparatus that would convert mechanical energy into heat by spinning a paddle in water. He observed that the loss of mechanical energy was proportional to the increase in temperature of the water.

(130)1cal=4.186J

Specific Heat

Specific heat is the heat required to raise 1kg of a substance by 1C.

Given a substance with a specific heat of c, the heat energy required to raise mass m of the substance by ΔT (change in temperature) is Q.

(131)Q=mcΔTc=QmΔT

The specific heat of water is 4186JkgC.14

Conservation of Heat Energy: Calorimetry

A calorimeter is an insulated container so that heat cannot go in or out. However, heat can flow between materials and objects within the calorimeter.

If not in equilibrium, cold objects heat up and hot objects cool down until thermal equilibrium is reached.

(132)Qgained=Qlost

The heat gained equals the heat lost.

Latent Heat and Phase Changes

A substance usually undergoes a temperature change when heat is transferred, but not always. When the substance is changing from one state of matter to another, it either absorbs or emits heat, but the temperature doesn't change while the phase change is happening.

Here are the types of phase changes:

Phase Changes Diagram

The following diagram shows the temperature of a substance as a function of heat added:

Temperature vs. Heat Added Graph

Notice how, in the flat segments, heat is being added or removed, but the temperature is not changing. The substance is going through a phase change.

When temperature is changing, Q=mcΔT applies. During a phase change, however, we must use the following equation:

(133)Q=mL

L is the latent heat, which is a constant, similar to the specific heat of a substance, except no temperature change is required. Latent heat is the heat per kilogram (Jkg) that must be added or removed when a substance changes from one phase to another. There are difference types of latent heats, depending on what kind of phase change a substance is going through.

Also, note that the energy in the vaporization/condensation phase change is much greater than the energy in the melting/freezing phase change.15

Transfer of Heat

There are 3 ways in which heat can be transferred.

  1. Conduction The process in which heat is transferred directly through materials in contact. The molecules/atoms in the material vibrate and pass some of their energy onto other particles. Thermal conductors conduct heat well (usually metals) while thermal insulators conduct heat poorly.

  2. Convection The process in which heat is carried from place to place by the bulk movement of a fluid. When part of a fluid is warmed, such as the air above a fire, the volume of the fluid expands and moves upward, and colder surrounding fluid replaces it. This flow is called a convection current. Natural convection results from differences in density, such as the air around a fire. Forced convection results from a forced external movement such as one caused by a fan or pump.

  3. Radiation The process in which heat energy is transferred through the means of electromagnetic waves. All objects continuously radiate energy. Generally, an object does not emit much visible light until the temperature of the object exceeds 1000K. The absorption of electromagnetic waves is just as important as the emission.

12: The Laws of Thermodynamics

The Basics of Thermodynamics

Thermodynamics is the branch of physics that is built upon the fundamental laws of nature that heat and work obey.

In thermodynamics, the collections of objects upon which attention is being focused is called the system while everything else is called the surroundings.

Walls that allow heat to flow are called diathermal and ones that do not are called adiabatic.

Heat and Internal Energy

Internal energy refers to all of the energy belonging to a system while it is stationary (no kinetic energy), including heat, nuclear, chemical, and strain (stretch or compression) energy.

Thermal energy is the portion of internal energy that changes when the temperature of a system changes.

Heat transfer is a process caused by a temperature difference between a system and its surroundings.

Work and Heat

Consider an ideal gas contained in a cylinder with a movable piston on top, as the following diagram shows:

Gas in Cylinder with Moveable Piston

In equilibrium, the gas has a volume of V and exerts a uniform pressure P on the walls and piston.

If the piston has a cross-sectional area of A, then the force exerted on the gas by the piston is F=PA.

If the piston is pushed down a distance of Δy by an external force of F, then work done on the gas is W=FΔy=PAΔy, and, because V=AΔy, we obtain:

(134)W=PΔV

All this is saying is that the work done on the gas is equal to the product of the pressure of the gas and the volume that it is changed.

This equation strictly represents the work done ON the gas and is highly dependent on the direction it is compressed.

The above equation only works if the pressure of the gas remains constant. When it does, it is called an isobaric process.

The First Law of Thermodynamics

In the past, we've only considered mechanical energy. However, we will now include internal energy.

Energy can be transferred between a system and its surroundings in two ways:

Suppose that a system gains heat and does no work. The internal energy, denoted by U, increases from Ui to Uf, the change being ΔU=UfUi. The heat, Q, positive when the system absorbs heat and negative when is releases heat.

Similarly, suppose that a system does W work on its surroundings, then ΔU=W. The work is positive when it is done on the system and negative when it is done on the surroundings.

The first law of thermodynamics states that, if a system undergoes a change from an initial state to a final state, where Q is the energy transferred to the system by heat and W is the work done on the system, the change in the internal energy of the system, ΔU, is given by:

(135)ΔU=UfUi=Q+W

Heat Engines and The Second Law of Thermodynamics

The second law of thermodynamics states that heat flows spontaneously from higher temperature to lower temperature.

A heat engine is any device that uses heat to perform work, as shown in the following diagram:

Heat Engine Diagram

Here's how it works:

Efficiency

An important characteristic of an engine is its efficiency. An engine that converts most of the input heat into work is efficient. Efficiency can be measured using the ratio between work produced and inputted heat. Efficiency is denoted by η.

(136)η=WQH

Efficiency is often measured in percent.

In an engine, the input heat, QH, is converted into work, W, and the left over heat, QC, is rejected into the cold reservoir. If there are no other energy losses in the engine, then:

(137)QH=W+QCW=QHQC

The work the engine provides can be expressed by difference in heat between the hot and cold reservoirs.

This can also be substituted into the efficiency equation to get:

(138)η=QHQCQH

Reversible and Irreversible Processes

A reversible process is one that can be performed so that, at its conclusion, both the system and surroundings have been returned exactly to their initial conditions, including energy. An irreversible process is one that does not satisfy these requirements.16

The Carnot Engine

Sadi Carnot proposed that a heat engine has a maximum efficiency when the processes within the engine are reversible. This is known as Carnot's Principle.

Carnot's Principle: No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficient than a reversible engine operating between then same temperatures. Furthermore, all reversible engines operating between the same temperatures have the same efficiency.

To describe the Carnot cycle, we assume that the substance working between the input temperature, Tc, and the output temperature, Th, is an ideal gas contained in a cylinder with a movable piston at one end.

Carnot Engine Diagram

This diagram illustrates the four phases of the Carnot cycle.

For information about PV diagrams and more on the work-heat-energy relationship of gasses, watch NOTES SMAY CH 12 Day 2 Finish 1st Law and Carnot Conceptual VIDEO starting from 23:59.

The difference in Qh and Qc is the work that the engine does. No, this is not a perpetual motion machine. There always needs to be a heat supply to account for the difference.

Carnot showed that the thermal efficiency of a Carnot engine is the following, where T is in Kelvin.

(139)ηc=ThTcTh

To maximize efficiency, Tc must approach absolute zero. This way, all of the input heat goes into doing work, and none of it is expelled. However, it is not possible to reach absolute zero. Therefore, the temperature of the hot reservoir (the input heat) can also be increased in an attempt to maximize efficiency.

Entropy and Disorder

In the 1800s, Rudolf Clausius created a concept called entropy, which is a measure of how disordered the universe is. "Disorder" is a measure of the usable energy in the universe and how spread out all of the energy is. A disordered arrangement is much more probable than an ordered one. Isolated systems tend toward greater disorder where everything is spread out.

Think about a block of ice that you took out of the freezer. For all the cold to be concentrated in the ice at any given moment in time is a very unlikely scenario, and, therefore, the heat starts spreading out until the ice melts and everything is the same. Unlikely scenarios tend toward likely and predictable ones.

You can, however, create processes that seem like they decrease entropy. For example, using the AC. The AC makes a cool house cooler, making the system more ordered. The reason this is possible is that, even though that specific action causes entropy to decrease, the processes that make the AC work (power lines, generators, power plants, etc.) increase entropy by more than the AC decreased it. Overall, there is still an increase in entropy.

An interesting way to think about entropy is using the Sun as an example. What do we get from the Sun? The obvious answer to this question is energy. However, the Earth also expels the same amount of energy back into space that it gets from the Sun. If it didn't, then the atmosphere would heat up and life wouldn't be possible. So knowing that the net energy change the sun provides is zero, what do we actually get from the sun?

The answer to this question is that we get low-entropy energy or ordered energy – usable energy that we then convert to high entropy energy that is no longer useful.

Every process that ever happens in the universe causes the total entropy of the universe to increase. This means that entropy is always increasing. The only process that doesn't cause entropy to increase are reversible processes, which are not really possible.

As current theories predict, in the extremely far future, after an unimaginable amount of time has passed, everything in universe will be so spread out that nothing interesting will happen again, and entropy will reach a maximum. This is known as the heat death of the universe.

Entropy is denoted by S.

Consider a reversible process. Qr represents the heat absorbed or expelled and the change in entropy by ΔS. The change in entropy is equal to the heat transferred divided by the absolute temperature (in Kelvin) of the system.

(140)ΔS=QrT

It can also be written as:

(141)ΔS=(QrT)f(QrT)i

When an irreversible process occurs and the entropy of the universe increases, the energy available for doing work decreases.

13: Vibrations and Waves

There are four types of waves (periodic motion):

Seismic waves are the only ones we will not be working with in this course.

Hooke's Law

Remember the equation that models the force of a spring: Fs=kx

k represents the proportionality constant that dictates the relationship between the distance you stretch the spring and the force with which the spring pulls back with. Stiff springs have large k values, whereas soft springs have small k values. The negative sign just means that the force is in the opposite direction as the displacement of the object. Fs is also called the restoring force.

Simple Harmonic Motion and Hooke's Law - SHM occurs when the net force along the direction of motion obeys Hooke's Law: the net force is proportional to the displacement from the equilibrium point and is always directed toward the equilibrium point.

General Waves Information

Furthermore, the acceleration of the target object in SHM can be solved for using a combination of Hooke's Law (F=kx) and Newton's Second Law (F=ma).

(142)ma=kxa=kmx

Elastic Potential Energy

Remember the equation for elastic potential energy. Elastic potential energy is also defined as the amount of work a spring is capable of doing. Remember that W=Fx. However, in the case of a spring, the force is dependent on the distance. F is a function of x, so to calculate the total capable work, we can take the following integral (we can omit the negative from F=kx because the direction does not matter when considering the work being done):

(143)PEs=W=Fdx=kxdx

Solving the integral, we get:

(144)PEs=12kx2

Velocity as a Function of Position

Considering a spring with an object stretched to length A, the total spring potential energy is 12kA2. During the motion of the spring and object, the energy will be a mix of elastic and kinetic energy. In total, they must add up to the original spring potential energy due to the laws of conservation of energy:

(145)12kA2=12mv2+12kx2

Solving the equation for v, we get:

(146)v=±km(A2x2)

This equation is only useful because it will be used in a future derivation. Just know conservation of energy, and specifically:

During the motion of the spring and object, the energy will be a mix of elastic and kinetic energy.

Comparing Simple Harmonic Motion with Uniform Circular Motion

Picture a turntable with a ball on the edge. As the turntable rotates, if we shine a light downward, the ball moves with constant velocity, and the shadow oscillates back and forth with simple harmonic motion. How do we know for sure, though?

Remember that v=±km(A2x2). Because km is a constant, we can rewrite this as:

(147)v=C(A2x2)

Look at the following diagram:

Comparing Simple Harmonic Motion with Uniform Circular Motion

Look at the two similar triangles, we know that

(148)sinθ=vv0

and

(149)sinθ=A2x2A

Because these are equivalent expressions, we can set them equal to each other and obtain:

(150)vv0=A2x2Av=v0AA2x2v=CA2x2

The x component of the position of the ball can be modeled using a cos graph. This means that sinusoidal motion is simple harmonic motion.

Period and Frequency

Period (T) represents how long it takes an object to complete one full revolution. The circumference is the distance traveled, so the ball moves a distance of 2πA in time T.

We can use these values to express the velocity:

(151)v0=2πAT

And also the time/period:

(152)T=2πAv0

Consider a quarter cycle, where the energy in the cosine wave turns from spring potential energy into kinetic energy: 12kA2=12mv02. This equation can be solved for Av0.

(153)Av0=mk

Substituting this into the period equation, we get:

(154)T=2πmk

Notice that the period does not depend on A.

The angular frequency, ω, is 2πT, which is:

(155)ω=km

Position, Velocity, and Acceleration as a Function of Time

This type of movement is known as rectilinear motion. Picture a sin wave. The x axis is time, and the y axis is position. Our position starts at 1 away from the origin.

The default equation has an amplitude of 1 and a period of 2π, as seen by:

(156)x=cosθ

The amplitude can be modified using a leading coefficient:

(157)x=Acosθ

θ is just the angular distance, so we can epxress is in terms of time as θ=ωt=2πft. Plugging this into the equation we get:

(158)x=Acos(2πft)

Because this is a function of time, we can find the first and second derivatives with respect to t to obtain equations for velocity and acceleration.

(159)v=Aωsin(2πft)
(160)a=Aω2cos(2πft)

Notice how these equations relate to period and angular frequency.

Think about a spring that has a mass m attached to it. At any point in time, the following relationship is true because they both equal F:

(161)ma=kx

isolating a and rearranging, we get:

(162)a=(km)x

Remember that x=Acos(2πft). Substituting this in, we get:

(163)a=A(km)cos(2πft)

Notice the similarities between this and:

(164)a=Aω2cos(2πft)

Looking at this form, we know that the following relationship it true:

(165)ω2=km

We can now obtain our angular frequency formula:

(166)ω=km

And because we know that ω=2πf, we can get our period formula, too:

(167)2πf=kmT=2πmk

Motion of a Pendulum

Pendulums oscillate back and forth, too, similar to SHM, but is it actually SHM?

Consider a pendulum with a length L that swung θ away from the vertical. The arc length is s=Lθ. The restoring force is proportional to the weight of the object, meaning that F=mgsinθ. Replacing theta with arc length, we get:

(168)F=mgsin(sL)

This equation is not in the form F=kx, so pendulums do not follow SHM.

However, for angles smaller than about 15, sinθθ. Which means that for pendulums that do not swing out very wide, we can say that they do follow SHM:

(169)F=(mgL)s

Going back to out period equation, substituting mgL for k gets us:

(170)T=2πLg

The describes the period of a pendulum.

Damped Oscillations

Based off of our equations, turntables would turn forever, and pendulums would swing indefinitely. However, in reality, friction and air resistance causes these contraptions to stop oscillating over time.

Waves are always caused by some sort of vibrating object. The mechanical waves discussed in this chapter require:

  1. A disturbance

  2. A medium that can be disturbed

  3. A physical connection or mechanism through which adjacent portions of the medium can influence each other

Types of Waves

Frequency, Amplitude, and Wavelength

λ represents the wavelength - the distance between two successive points that behave identically. We can use this distance to define the velocity of a wave: v=xt=λT:

(171)v=λf

Wave Speed Under Tension

Wave speed depends on the tension in the string. If a string under tension is pulled sideways and released, the tension, F​, is responsible for accelerating a particular segment back to its original position.

(172)vF

The wave speed is inversely dependent on the mass per unit length, μ​, of the string. Think about a guitar. This basically means that the thicker the string (the more the mass), the slower the wave and the deeper the sound.

(173)v1μ

The exact relationship is:

(174)v=Fμ

This means that the wave speed depends on only the properties of the string, not the amplitude.

If you're curious as to how exactly you derive this formula, check out this YouTube video.

Superposition and Interference

Two traveling waves can meet and pass through each other without being altered or destroyed.

The superposition principle states that when two or more traveling waves encounter each other while moving through a medium, the resultant wave if found by adding together the displacements of the individual waves point by point.

Reflection of Waves

When a wave encounters a boundary, like a wall, it is reflected. There are two possible scenarios that can be thought of using a rope:

  1. The end of the rope is fixed to the wall If the end is fixed, the wave will not only bounce back, but it will invert and flip over.

  2. The end of the rope is free to move (eg. the end is a loop tied around a pole) If the end is free to move, the wave will still bounce back, but it will not invert or flip.

15: Electric Forces and Electric Fields

Around 700 B.C., the Ancient Greeks conducted the earliest known study of electricity. They noticed that small objects would be attracted to amber if it was rubbed with wool.

These electric charges are caused by a difference of electrons. In a conductive object, such as metal. Electrons are allowed to pass freely among the particles that make up the metal. This is known as a "sea of electrons".

Electrons are negatively charged, and so there are two possible states that a material can have:

The most important law about these electrons is that like charges repel and unlike charges attract. Specifically, here is a table of all possible combinations (Ø means neutral):

Charge 1Charge 2Attraction
+-Attract
++Repel
--Repel
+ØAttract
-ØAttract
ØØNo attraction

Two objects attract when there is a difference in type of charge. Otherwise, they repel.

Properties of Electric Charges

In 1909, Robert Milikan proposed that if an object is charged, its charge is always a multiple of a fundamental unit of charge; it is quantized. This fundamental unit of charge is equivalent to the charge that one electron carries, denoted by e.

(175)1e=1.60219×1019C

C​ is the SI unit of this measurement, called coulomb.

Insulators and Conductors

In conductors, electric charges move freely in response to an electric force. All other materials are called insulators.

Charging by Conduction (Contact)

When a neutrally charged object comes into contact with a charged one, the amount of charge "equalizes" between the two objects. Meaning that if a neutral object comes into contact with a negatively charged object, it also becomes negatively charged. The object doing the charging remains negatively charged, but to a lesser degree. (Works the same for positive)

Charging by Induction

Charging by Induction Diagram

Look at the diagram above. The Metal sphere is originally neutrally charged. When the negatively charged ebonite rod is held near it, the electrons escape to as far away from the rod as possible because they repel, making the side closer to the rod positively charged. When a grounding wire is introduced, the electrons have an escape route and move even farther away from the negatively charged rod, leaving the sphere. If the grounding wire is removed, then the sphere remains positively charged.

Coulomb's Law

In 1785, Charles Coulomb established the fundamental law of electric force between two stationary charged particles.

This equation is analogous to the Universal Law of Gravitation.

The magnitude of the electric force, F, between charges q1 and q2 separated by a distance r is given by:

(176)F=ke|q1||q2|r2

ke is the Coulomb constant.

(177)ke=8.9875×109Nm2C2

This force is attractive if the charges have opposite signs and repulsive if the charges have the same sign.

Also, the electric forces between elementary particles are far stronger than the gravitational forces between them.

The Superposition Principle

When a number of separate charges act on the charge of interest, each exerts an electric force. These electric forces can all be computed separately, one at a time, then added as vectors.

Electric Fields

Remember that field forces are types of forces that don't need contact between objects: gravity and electricity. A person named Michael Faraday proposed the idea of an electric field. It is a field, analogous to a gravitational field that is said to exist around a charged object.

The definition of an electric field is the amount of force per coulomb of charge that would be exhibited on a positively charged particle at any given point.

Think back to gravity. A planet, for example, exhibits a gravitational field around it. Any object within the gravitational field is subject to a force per kilogram of mass just like for an electric field. Gravitational fields deal with the mass of objects, while electric fields deal with the charge of objects.

An electric field is analogous to the acceleration caused by gravity.

(178)E=Fq0

An electric field produced by a charge Q equals the force that it exerts on a hypothetical charge q0 divided by q0. The units of electric fields are NC.

Current topic equations will be on the left, while analogous gravity equations will be on the right.

This means that:

(179)F=q0EF=ma

Now, look at the two force law equations:

(180)F=keq0Qr2F=GmMr2

Substituting, we get:

(181)q0E=keq0Qr2ma=GmMr2(182)E=keQr2a=GMr2

The final electric field equation is:

(183)E=keQr2

All of these equations and ideas are exactly analogous to gravity.

Relationship Between Electric Fields and Distance:

(184)E=ΔVd

This is an alternate equation for electric fields.

Electric Field Lines

A convenient way to visualize electric field patterns is to draw lines pointing in the direction of the electric field vector at any point.

The magnitude of E is large where the lines are close and small when the lines are far apart.

Positive, Negative, and Neutral Particle Electric Field Lines Diagram

Rules For Drawing Electric Field Lines

  1. Lines begin on positively charged particles and end on negatively charged particles unless they approach infinity.

  2. The number of lines drawn leaving or ending a particle is proportional to the strength of the charge.

  3. Field lines cannot cross.

Examples of Electric Field Lines

Electric Field Lines Example

Conductors in Electrostatic Equilibrium

When no net motion of charge occurs within a conductor, the conductor is said to be in electrostatic equilibrium.

16: Electrical Energy and Capacitance

Potential Difference, Electric Potential, and Voltage

Electric potential or Volts is a property of space that describes how much electric potential energy could be present at a given point per unit of charge.

(185)1V=1J1C

Voltage is electric potential difference, represented by ΔV. Given two points with different electric potentials, the voltage can be expressed as follows:

(186)ΔV=ΔPEq=VBVA

For example: if point A is at 20V and point B is at 80V, then the voltage of the system is 60V.

The work done by an electric field is Fd where d is the distance it pushes charge. Because F=qE, we can say the work done by the electric field is:

(187)W=qEd

As the electric field does work, ΔKE is positive because kinetic energy increases, and ΔPE is negative, because potential energy decreases. This means that:

(188)W=ΔKE=ΔPE

Putting everything together we obtain:

(189)ΔPE=Wby field=qEd=qΔV

It is also important to note, units-wise, that 1NC=1Vm.

Electric Potential Energy

In electric circuits, points at which the potential is zero is often defined by grounding some point in the circuit. For a point charge, however, similar to gravity, a point of zero potential can be defined as being an infinite distance away from the charge. The same goes for electric potential energy. Electric potential energy can be derived as follows:

Electric potential energy can be defined as the work needed to move a positive particle toward another positive particle with an electric field E from infinitely far away (where PE is 0) to a distance r away. This resembles gravitational potential energy very closely. If you remember, GPE was negative, because gravity naturally attracts all objects. However, in the case of electric fields, positives repel, so the final result should be positive.

Electric Potential Energy Diagram

Because the direction of force needed is to the right, and the electric field pushes the other way, the the force by the electric field is qE.

(190)PE=W=rFdx

Because F=qE in this case and x=r, we can say:

(191)PE=rqEdr

Remember the formula: E=keQr2

(192)PE=rkeqQr2dr

Solving this integral, we get PE=keqQr. Using standardized variables:

(193)PE=keq1q2r

Electric Potential

Remember that the electric potential is ΔV=ΔPEq. This also means that:

(194)V=PEq

Going back to the electric potential energy formula and plugging it in, we get:

(195)V=keqQrq

Simplifying and converting to standardized variables, we get the formula for electric potential:

(196)V=keqr

Potentials and Charged Conductors

The Electron-Volt

One electron-volt, 1eV, is defined as the energy that an electron (or proton) gains when accelerated through a potential difference of 1V.

(197)1eV=1.60×1019J

Equipotential Surfaces

An equipotential surface is one on which all points are at the same potential. No work is required to move a charge at a constant speed on an equipotential surface. Equipotential lines (isolines) are always perpendicular to electric field lines.

Equipotential lines and Electric Field Lines

Capacitance and Parallel-Plate Capacitors

A capacitor can be charged in order to store energy in circuits that can later be reclaimed for use. A diagram of a typical capacitor is shown below.

Capacitance Diagram

The insulator in the middle (usually air) gets charged in a way where electricity can be released from the capacitor at a later time. Current does not pass through a capacitor. The charge on the one plate induces the charge on the other plate.

The capacitance, C, of a capacitor is the ratio of the magnitude of the charge on either conductor (plate) to the magnitude of the potential difference (voltage) between the conductors.

(198)C=QΔV

The SI unit of capacitance is the farad (F​).

The capacitor diagram shown above is what is known as the parallel plate capacitor. The electric diagram version of the one shown above is (the short side of the battery is the negative side):

Capacitor Electrical Diagram

For a parallel plate capacitor:

Two constants are also involved:

Putting everything from above together, we get the following equation for capacitance:

(200)C=ϵ0Ad

It is important to note that the electric field in between the plates is nearly constant.

Capacitors in Parallel

Two capacitors in parallel in a circuit are arranged as follows.

Capacitors in Parallel Diagram

Capacitors in parallel both have the same voltage across them. The charges of the capacitors, however, add up: Q=Q1+Q2​. The two capacitors can be replaced with a single capacitor of charge Q. This capacitor, Ceq is: Q=CeqΔV. Because the charges add up, we can say: CeqΔV=C1ΔV+C2ΔV. The voltages cancel, meaning we can conclude the following:

For Capacitors in Parallel:

For n capacitors in parallel, the equivalent single capacitance is:

(201)Ceq=k=1nCk=C1+C2+...+Cn

The equivalent capacitance of a parallel combination of capacitors in greater than any of the individual capacitances. This is true because capacitors in parallel is like increasing the area of a single capacitor.

Capacitors in Series

Two capacitors in series are arranged as follows:

Capacitors in Series Diagram

Capacitors in series both have the same charge across them. The voltages of the capacitors, however, add up: ΔV=ΔV1+ΔV2. The two capacitors can be replaced with a single capacitor of charge Q. This capacitor, Ceq is: ΔV=QCeq. Because the voltages add up, we can say: QCeq=QC1+QC2​. The charges cancel, meaning we can conclude the following:

For Capacitors in Series:

  1. Charge is the same

  2. Voltages add up

For n​ capacitors in series, the equivalent single capacitance is:

(202)1Ceq=k=1n1Ck=1C1+1C2+...+1Cn

The equivalent capacitance of a series combination is always less than any individual capacitance in the combination. This is true because capacitors in series are like increasing the distance between plates.

Energy Stored in a Charged Capacitor

Capacitors can store electrical energy, and that energy is the same as the work required to move charge onto the plates. As soon as the capacitor's plates touch, or have have a conductor placed between them, the electricity will flow.

Because the units for V is JC and the units for Q is C, work can be obtained by multiplying them together. For a battery continually delivering charge to a capactior, the fromula looks like:

(203)E=W=0QVdQ=0QQCdQ

Solving this integral, we get that the energy, E, is:

(204)E=Q22C=C(ΔV)22

17: Current and Resistance

Electric Current

Current is the rate at which charge flows. Suppose ΔQ is the amount of charge that flows through an area A in a time interval Δt. Then the current, denoted by I, is given by:

(205)I=ΔQΔt

The SI unit of current is the Ampere (A).

Conventional current is defined by the direction that positive charges flow.

As charges move along a wire, electric potential is continually decreasing.

Current and Drift Speed

In a wire with a cross-sectional area of A, the volume of a length Δx is AΔx. If n represents the number charge carries per volume, then the total number of charge carries is given by nAΔx. If we want to find the total charge, we multiply by the charge that each charge carrier carries: ΔQ=nAΔxq.

Charge carries move with a constant average speed called the drift speed.

(206)vd=ΔxΔt

Substituting this into the previous equation, and dividing by Δt, we get the final formula for current:

(207)I=ΔQΔt=nqvdA

Resistance and Ohm's Law

When a voltage is applied across the ends of a metallic conductor, we find that the current is proportional to the voltage.

(208)IΔV

This can be rewritten as ΔV=IR where R is the proportionality constant, also known as the resistance of the conductor.

Resistance is defined as the ratio of voltage across the conductor to the current it carries:

(209)R=ΔVI

The SI unit for resistance is Ohms (​).

Resistivity

The resistance of a conductor increases with length, and decreases with area. Using a proportionality constant ρ, we can conclude the following:

(210)R=ρlA

This proportionality constant is called the resistivity of the material.

Temperature Variation of Resistance

For most metals, resistivity increases with increasing temperature. As the heat goes up, molecular motion also goes up and it becomes harder for an electron to smoothly make its way through the material. The following equations describe the linear relationship between resistivity/resistance and temperature (α is the temperature coefficient of resistivity):

(211)ρf=ρi(1+α(TfTi))
(212)Rf=Ri(1+α(TfTi))

Superconductors

Superconductors are a class of metals and compounds with resistances that fall virtually to zero below a temperature called the critical temperature. This temperature is usually extremely low.

Electrical Energy and Power

The chemical energy that's stored in a battery is continuously transformed into thermal energy (through collisions). These collisions are what cause a lightbulb to heat up and glow.

As a charge ΔQ passes through a resistor, it loses energy ΔQΔV. If it takes the charge Δt time to pass through the resistor, then the rate at which it loses energy is:

(213)ΔQΔVΔt=IΔV

The power, representing the rate at which energy is delivered to the resistor, is therefore:

(214)P=IΔV

This equation can be slightly modified using ΔV=IR to get other forms:

(215)P=IΔV=I2R=(ΔV)2R

18: Direct Current Circuits

EMF

EMF stands for electromotive force. It represents a voltage. In a battery, the EMF is the "theoretical" voltage. In reality, by the time the current comes out, it experiences some internal resistance in the battery. This is the "actual" voltage of the battery.

For example, a battery may be labeled as 9V. This is the EMF. When measuring, though, a reading of 8.9V may be observed. This is the voltage.

EMF is represented by ϵ​.

With a current I and internal resistance of r, the voltage of the battery can be written as follows:

(216)ΔV=ϵIr

Consider an external resistor with resistance R. The same current travels through, and the same voltage affects it. Meaning:

(217)IR=ϵIr

Soling for current:

(218)I=ϵR+r

We generally assume the internal resistance of a battery is zero unless otherwise specified.


Resistors in series and in parallel work very similar to capacitors in series and parallel, so I will not include the repetitive derivations.

Resistors in Series

Two resistors in series in a circuit are arraganged as follows:

Resistors in Series

For Resistors in Series:

For n resistors in series, the equivalent single resistance is:

(219)Req=k=1nRn=R1+R2+...+Rn

Resistors in Parallel

Two resistors in parallel in a circuit are arranged as follows:

Resistors in Parallel

For Resistors in Parallel:

For n resistors in parallel, the equivalent single resistance is:

(220)1Req=k=1n1Rn=1R1+1R2+...+1Rn

Kirchhoff's Rules and Complex DC Circuits

There are many way in which resistors can be connected so that the circuits formed cannot be reduced to a single equivalent resistor. Therefore, we use Kirchhoff's Rules:

Possible Loop Rule Cases:

  1. Loop in the same direction as battery (negative to positive) battery 1

  2. Loop in the opposite direction as battery (negative to positive) battery 2

  3. Loop in the same direction as resistor (current is to the right) resistor 1

  4. Loop in the opposite direction as resistor (current is to the right) resistor 2

RC Circuits

RC circuits are ones which has a resistor and a capacitor.

Let's consider DC circuits containing capacitors. In these circuits, current varies with time. As the capacitor becomes more charged, the rate at which it is able to receive charge (current) decreases.

The maximum charge capacity of a capacitor is:

(221)Qmax=Cϵ

ϵ is the maximum voltage across the capacitor. Once the capacitor is fully charged, I=0A.

If we assume that the capacitor is uncharged before the circuit is on, the charge on the capacitor at time t is given by the following equation.

Charge at time t of a capacitor being charged:

(222)q=Qmax(1etτ)

τ is the time constant:

(223)τ=RC

If we assume that the capacitor is charged with a charge Q and is being discharged, the charge on the capacitor at time t is given by the following equation.

Charge at time t of a capacitor being discharged:

(224)q=Qetτ

Household Circuits

Electric companies distribute electric power through a pair of wires. One of them is connected to the ground, and the other (called the hot wire) is at a potential of 120V. Because the ground has a potential of 0V, the potential difference, or voltage achieved is 120V.

Electrical devices in a home are connected in parallel.

A circuit breaker (or fuse) is connected in series with the wire entering the home. If too much is connected to the power source and the wires get too hot due to the current, the circuit breaker pops the fuse, disconnecting the entire circuit, and preventing a possible fire.

Some appliances require 240V. For this, the neutral wire is sometimes replaced with one that is kept at 120V​.

For safety, many plugs have a third wire, called a case ground. This is so that in case another wire fails, there is still a grounding wire. If there wasn't, something else must act at the ground wire (a person).

The Measurement of Current and Voltage

Current can be measure with ammeters.

Voltage can be measure with voltmeters.

Electricity Review Material

There are a lot of different important equations regarding electricity. Here is a PDF document with all the electricity equations you need to memorize in one. Click the link above or the image below.

19: Magnetism

There are 2 types of magnetic materials:

The direction of a magnetic field at any location is the direction in which the north pole of a compass would point.

Magnetic Fields

When a charged particle is moving through a magnetic field, a magnetic force acts on it. The particle must be moving for there to be a force. The equation for magnetic force is the following (B​ is the magnetic field):

(225)F=qvBsinθ

θ is the angle between v and B.

The SI units of B are T (tesla). Another form of units for magnetic fields is G (gauss), where 1T=104G​.

The direction of the magnetic force is always perpendicular to both v and B.

The Right-Hand Rule:

Right-Hand Rule

The right-hand rule can be used to determine the directions of everything. Everything shown above has right angles between them.

Direction of Magnetic Fields:

Magnetic Field Directions

Magnetic Force on a Current-Carrying Conductor

Magnetic forces are also exerted on a current-carrying wire, similarly to a single charged particle.

Let's take a wire with length l and cross-sectional area A. Let n represent the number of charges per unit volume.

This means that the total force would be represented by:

(226)Fmax=qvdBnAl

Remember that I=nqvdA. Plugging this in, we get:

(227)F=BIlsinθ

Torque on a Current Loop and Electric Motors

If a current travels through a loop and is placed inside a magnetic field, a torque will be exerted on the loop as the result of a magnetic force.

Torque on a Current Loop Side View

In the diagram above, the forces on side a are zero, but the forces on side b are BIb each. Looking from the top:

Torque on a current Loop Top View

When calculated, the total torque is:

(228)τ=BIA

A is the area: A=ab.

This only works if everything is perpendicular. Let θ be the angle between the magnetic field and the line perpendicular to the plane of the loop. Notice how in the previous diagram, this value is 90.

The torque is also dependent on the amount of times the wire is coiled. This number is denoted by N.

For short, the expression IAN is sometimes written as μ, which a vector called the magnetic momentum of the coil.

(229)τ=BIANsinθ=μBsinθμ=IAN

Motion of a Charged Particle in a Magnetic Field

Consider a positively charged particle moving through a magnetic field in a way such that the velocity of the particle is always perpendicular to the magnetic field. Since the force is perpendicular to both the magnetic field and the velocity, it acts as a centripetal force.

(230)F=qvB=mv2r

Magnetic Field of a Long, Straight Wire and Ampere's Law

In 1819, Oersted found that en electric current in a wire caused there to be an electric field around the wire. The magnetic field goes in a circle around the wire. Using another right hand rule, it is possible to figure out the direction of the magnetic field.

Point your thumb in the direction of the current through the wire. The direction that the rest of your fingers curl is the direction of the magnetic field around the wire.

The direction of the magnetic field at a given point is tangent to this circle.

B is proportional to the current and inversely proportional to the distance. The relationship has a proportionality constant, μ0, the permeability of free space.

(231)B=μ0I2πrμ0=4π×107TmA

Magnetic Force Due to Wires With Current

Force on a Particle When acting on a particle, the force is:

(232)F=qvμ0I2πr

Force on another Wire The force that a wire exerts on another wire is:

(233)F=μ0I12πrI2l

For two wires with current, they both exert magnetic forces on each other mutually and symmetrically.

Magnetic Domains

Due to the spins of electrons within materials, some materials, like metals, have electrons that don't have a pair counteracting the spin. This motion causes there to be a small magnetic field. However, each particle in a material usually has magnetic fields oriented in completely random directions, meaning that the overall magnetic field is near zero.

When put under a magnetic field, though, all the orientations can align, magnetizing the material.

A ferromagnetic material is one which has enough lone electrons within its atoms to be magnetized.

A magnetic domain is a large group of atoms with spins that are aligned.

20: Induced Voltages and Inductance

In 1819, Hans Oersted discovered that an electric current exerted a force on a magnetic compass. Electric current can produce a magnetic field and vice versa.

Induced EMF and Magnetic Flux

Using an experiment, Michael Faraday figured out that a current can be created using a changing magnetic field. To quantify how a magnetic field changes, we use concept called magnetic flux.

The magnetic flux ΦB through a loop of wire with area A is:

(234)ΦB=BA=BAcosθ

B is the component of B perpendicular to the plane of the loop. θ is the angle between B and the normal to the plane of the loop. The SI unit of magnetic flux is the weber (Wb​).

Magnetic flux can be visualized by thinking of, at a certain angle, how many magnetic field lines can pass through a coil of wire.

Faraday's Law of Induction

Electric current is induced when there is a change in magnetic flux. In turn, an EMF is produced, which is described by:

(235)ϵ=NΔΦBΔt

The negative sign simply indicated the polarity of the induced emf and, therefore, the direction of the induced current. Since it is hard to visualize the direction from the equation, when calculating, the negative can be omitted to obtain the magnitude of the emf, and Lenz's Law can be used to find the direction.

Lenz's Law When there is a change in magnetic flux, a direction can be associated with it: through the loop forward or backward. The direction of current is in the direction that opposes the change in flux. Use the right hand rule to find the direction of current from the direction of opposing flux.

Motional EMF

When a conductor moves through a magnetic field, an EMF is produced. Think of a metal bar moving to the right through a magnetic field directed into the page. Using the right hand rule, the motion of negative charges within the bar is downward. This means that a current is induced upward.

If the circuit is completed, this concept is analogous to increasing or decreasing the area of the closed circuit loop.

Remember that ΔΦB=BΔA. With a bar of length l moving in a closed loop, the change in area is BlΔx, where Δx is the distance the bar moves. Putting this into the induced voltage equation, we get:

(236)|ϵ|=ΔΦBΔt=BlΔxΔt=Blv

21: Electromagnetic Waves

Electromagnetic waves are composed of fluctuating electric and magnetic waves. All light is electromagnetic waves.

The Transformer

A transformer is used to either upscale or downscale a voltage.

A transformer works by magnetizing a conductive loop with a certain amount of coils. On the other side of the loop, the change in flux induces a different voltage. To always make sure the flux is changing, transformers only work with AC current.

Transformers Diagram

We know that the induced voltage is ΔV1=N1ΔΦBΔt and the induced voltage on the other side is ΔV2=N2ΔΦBΔt. The change in flux and the time is identical, so we can arrange the two equations to get the following relationship.

(237)ΔV1N1=ΔV2N2ΔV2=N2N1ΔV1

N2N1 is the ratio.

At first, this may seem like free energy, but, in fact, the power remains the same. This means that as the voltage varies, current also varies: P1=P2

(238)I1ΔV1=I2ΔV2

Maxwell's Predictions

During the early stages of study, electric and magnetic fields were thought to be unrelated. In 1865, James Clerk Maxwell provided a mathematical theory that showed a close relationship between all electric and magnetic phenomena:

  1. Electric field lines originate on positive charges and terminate on negative charges (free charges exist).

  2. Magnetic field lines always form closed loops (free charges do not exist).

  3. A varying magnetic field induces an emf and hence an electric field.

  4. Magnetic fields are generated by moving charges.

Electromagnetic waves are waves in which electric and magnetic fields traverse space in sync.

Production of Electromagnetic Waves by an Antenna

In an antenna, if AC current is running through it, the circuit loses energy in the form of electromagnetic waves. In fact, any circuit with AC current radiates EM waves.

As the charges move back and forth, an electric field parallel to the antenna is produced and radiates outwards. Additionally, a magnetic field perpendicular to the antenna is also produced.

Properties of Electromagnetic Waves

Plane wave - a transverse wave that oscillates in one plane.

Electromagnetic Wave Diagram

  1. Electromagnetic waves travel at the speed of light.

  2. Electromagnetic waves are transverse waves.

  3. The ratio of the electric field to the magnetic field in an electromagnetic waves is the speed of light.

    (239)EB=c
  4. Electromagnetic waves carry both energy and momentum.

The Spectrum of Electromagnetic Waves

Because all electromagnetic waves travel through space at the speed of light, their wavelength and frequency are related by:

(240)c=fλ

Electromagnetic Wave Spectrum

This is the spectrum of EM waves.

Important unit notes:

  1. 1 micrometer=1µm=106m

  2. 1 nanometer=1nm=109m

  3. 1 angstrom=1Å=1010m​​

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22: Reflection and Refraction of Light

Light has a dual nature. In some cases, it acts like a wave, in others, it acts like a particle.

The Nature of Light

Newton explained light as a stream of particles. Huygens showed that a wave theory of light could also explain the laws of reflection and refraction. At the time, all waves needed a medium to travel through, and, if light were a wave, it would bend around corners (light does actually bend around objects (diffraction)).

Over the course of many experiments, it was concluded that light acts both as a wave and as a particle.

The energy of a photon of light is proportional to its frequency.

(241)E=hf=hcλPlanck’s constant: h=6.63×1034Js

Reflection of Light

When light traveling in a transparent medium encounters a boundary, part of the incident ray is reflected.

Types of Reflection

The law of reflection states that the reflected ray has the same angle to the normal with which the incident ray hits the boundary.

(242)θ1=θ2

Refraction of Light

When light traveling through a transparent medium encounters another transparent medium, part of the ray is refracted and enters the second medium.

Refraction of Light

The angle of refraction, θ2, has the following relationship to the velocity of the light.

(243)sinθ1sinθ2=v1v2=constant

The index of refraction, n, of a medium is defined as:

(244)n=cv

As light travels from one medium to another, its frequency must stay the same, and, using the formula v=fλ and combining two indices of refraction, we get:

(245)λ1n1=λ2n2

Combining this with the previous equation, we obtain Snell's Law of Refraction:

(246)n1sinθ1=n2sinθ2

Dispersion and Prisms

The index of refraction of a material actually depends on the wavelength of light entering it. This is called dispersion. The index of refraction decreases with increasing wavelength. This means that the longer the wavelength of light, the less it refracts.

A prism uses this to separate light into its respective colors that make it up.

In a prism, the angle between the entering and leaving waves is called the angle of deviation, denoted by δ.

Angle of Deviation

Total Internal Reflection

When light enters a medium in which it bends away from the normal, at some angle called the critical angle, the refracted beam will travel along the surface of the material. At every angle beyond this, the light will completely reflect back inside of the material. This is called total internal reflection. This angle can be found using Snell's Law: n1sinθc=n2sin90

(247)sinθc=n2n1,n1>n2

23: Mirrors and Lenses

Flat Mirrors

On object placed in front of a mirror is at point O. The distance from O to the mirror is the object distance, denoted by p. The image on the other side of the mirror, the virtual image is a distance q from the mirror.

Flat Mirrors

Images are formed at the point where rays of light intersect.

Images have three qualities:

Images Formed by Spherical Mirrors (Concave Mirrors)

The center of curvature of a spherical mirror is denoted by c. The distance from c to the mirror (the radius) is called the principle. There are two possible cases of a reflection:

When the light rays come in close to parallel, the reflected light rays can converge, however, if they are not close to parallel, they do not converge, forming blurry image known as a spherical aberration.

The magnification is:

(249)M=hh=qp

The mirror equation, where R is the radius, is:

(250)1p+1q=2R

This can be rewritten as:

(251)1p+1q=1f

f is the focal length, where:

(252)f=R2

The focal length represents where the image of an object that is infinitely far away appears.

Mirrors

Ray Diagrams for Mirrors

In ray diagrams, there need to be at least two (preferably three) rays drawn.

  1. The first ray must come in parallel and pass through the focal point.

  2. The second ray must pass through the focal point and exit parallel.

  3. The third ray must reflect back on itself.

A real image is formed where the rays converge. If they seemingly don't converge, by extending the rays, a hypothetical convergence point can be found. This represents a virtual image being formed.

Atmospheric Refraction

The Sun appears to be above the horizon even after it has actually set. This is because air does have a slight index of refraction, meaning it bends light toward the ground in the case of the Sun, slightly.

A mirage is formed by hot air. This can be seen as the "shimmer" above a long stretch of road on a hot day. As the road warms up, so does the air around it, and warm air has a different index of refraction. The light from an object traveling toward the ground can bend upward and can be seen as a "reflection".

Thin Lenses

Lenses are two-sided pieces of glass that bend light in certain ways. The two sides can consist of convex, concave, or straight edges.

Converging lenses are thickest in the middle.

Diverging lenses are thickest at the edges.

The equations for lenses are the same as those for mirrors.

The focal length for a lens in air is related to the curvatures of its front and back surfaces and to the index of refraction n of the lens material by:

(253)1f=(n1)(1R11R2)

Lenses

Combination of Thin Lenses

When multiple lenses are used one after the other, the image formed by the first lens is treated as the object for the second lens.

Aberrations

Lenses and mirrors aren't perfect, so the light may not converge exactly at a point, sometimes producing a blurry-looking image. This is called a spherical aberration.

Similarly, different color light has different wavelengths, meaning that they get refracted in slightly different ways. This is called a chromatic aberration. The image on the left was captured with a camera that does not correct chromatic aberrations, and the image on the right has the aberration corrected. Notice the color shifts near edges.

Chromatic Aberration

24: Wave Optics

Geometric optics, covered by the last chapter, depends on the particle nature of light. Wave optics, however, depend on the wave nature of light.

Conditions for Interference

2 waves can add together constructively or destructively. For interference between two sources of light to be observed, the following conditions must be met:

  1. The sources must be coherent, which means the waves they emit must maintain a constant phase with respect to each other.

  2. The waves must have identical wavelengths.

You can also hear interference. This is how noise-cancelling headphones work. They emit a sound which attempts to cancel out background noise by interfering with it destructively.

Young's Double Slit Experiment

This is what the double slit experiment looked like:

Double Slit Experiment

The visible patters consists of bright and dark parallel bands called fringes.

Let's say the two slits are separated by a distance d, and the distance between the slits and the screen is L​. The wave emitted from one slit may travel a difference distance than the wave emitted from the other slit to a point on the screen. This difference in distance depends on the angle between the slits and the point on the screen.

For the following two equations, m is an integer multiple, called the order number.

Constructive Interference:

(254)δ=dsinθbright=mλ

Destructive Interference:

(255)δ=dsinθdark=(m+12)λ

The distance between the fringes can also be obtained.

If L is the direct distance from the slits to the screen and d is the distance between the slits, then:

Bright Fringes:

(256)y=λLdm

Dark Fringes:

(257)y=λLd(m+12)

Phase Change due to Reflection

When a light wave reflects from a surface the two possible cases are possible.

  1. If the n-value of the reflecting surface is less than the surroundings, then the phase stays the same.

  2. If the n-value of the reflecting surface is greater than the surroundings, then the phase shifts by 180​.

In summary, an electromagnetic wave undergoes a phase change of 180​ upon reflection from a medium that has an index of refraction higher than the one in which the wave was traveling.

Interference in Thin Films

Interference in Thin Films Diagram

The diagram above shows how interference can happen in films. Depending on whether or not the film has a higher or lower index of refraction compared to the mediums either above or below it, the light in either point may reflect with or without a phase change.

If the thickness of the film is t, then:

The equation for constructive interference is:

(258)2nt=(m+12)λ

The equation for destructive interference is:

(259)2nt=mλ

Newton's Rings

When the thickness of a film varies, it can result in interesting patters, such as rings:

Newton's Rings Photo

Diffraction

Huygen's Principle - Every point on a wave form acts as a source of tiny wavelets that move forward with the same speed as the wave.

Huygen's Principle means that, when light passes through any opening similar to the size of the wavelength of the light itself, an interference pattern emerges.

Single-Slit Diffraction

This is the phenomenon mentioned above. Because every point of the light's path acts as a source of waves, light from one portion of the slit can interfere with light from another portion.

The equation for destructive interference for single-slit diffraction is:

(260)sinθ=mλa

a is the width of the slit.

Additionally, the points of constructive interference occur halfway between points of destructive interference.

Single Slit vs. Double Slit Interference Patterns

The resulting interference patterns from single-slits and double-slits are very similar. However, there are some key differences.

  1. Width of the 0th order bright fringe In a double-slit interference pattern, the width of the 0th order bright fringe is the same as the width of any other bright fringe. However, in a single-slit interference pattern, the width of the 0th order bright fringe is twice that of all the other bright fringes.

  2. Rate of intensity with respect to distance The rate at which the intensity of light decreases as you move away from the center is greater for a single-slit interference pattern than it is for a double-slit interference pattern. In other words, the intensity of the bright fringes gets weaker faster for single-slit interference patterns than it does for double-slit interference patterns.

Diffraction Grating

The diffraction grating is a device that analyzes a light source by using a large number of equally spaced parallel slits. The condition for maxima in the interference pattern, with d being the spacing between slits and θ being the angle of the fringe, is:

(261)dsinθ=mλ

The interference pattern for a diffraction grating differs slightly compared to a double slit interference pattern, shown below:

Double Slit Interference Pattern Example:

Double Slit Interference Pattern Example

Diffraction Grating Interference Pattern Equivalent:

Diffraction Grating Interference Pattern Equivalent

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26: Special Relativity

Newtonian physics fails when considering particles moving close to the speed of light. The concepts of special relativity which explains this often violates our common sense about dimensions/time. It is, however, verified by particle accelerators all throughout the world.

The speed of any particle with mass must be less than the speed of light. Furthurmore, the concepts of force, momentum, and energy do not apply for rapidly moving objects. Observers moving at different speed will measure time and displacements differently.

The Principle of Galilean Relativity

In order to be able to describe a phsyical event, it's necessary to choose a frame of reference. The Principle of Galilean Relativity suggests that the laws of mechanics must be the same in all inertial frames of reference.

The Speed of Light

According to Einsein, the speed of light is the same for all observers.

If this is the case, though, then Galilean Relativity doesn't work. Imagine this: you're in a car that is moving at the speed of light and you turn your headlights on. What happens? If light were to appear, then it would be moving at twice the speed of light relative to an outside observer. In other words, there would be no constant speed of light.

The addition laws of velocity are therefore different.

Einstein's Principle of Relativity

In 1905, Einstein proposed a theory which solved the contradiction of the velocity addition. However, his proposal altered the way we think about space and time.

The two postulates of relativity:

  1. All the laws of physics are the same in all inertial frames.

  2. The speed of light in a vacuum is always 2.998×108ms​.

In relativistic mechanics, there is no such thing as absolute length or absolute time.

Time Dilation

Common sense indicates that time passes just as quickly for someone on the ground than someone moving in a space craft. In contrast, special relativity reveals that the person on the ground measures time passing more slowly than the person in the space craft.

Suppose there are two light clocks, as shown in the picture below:

Light Clocks

The light for the scenaria in which the object is moving seems to travel a longer distance in the same amount of time, but that cannot be. Therefore, time is dilated.

According to the astronaut in the spacecraft, the time interval Δt0 required for light to bounce up and down is 2Dc​.

(262)Δt0=2Dc

For the observer on the ground, however, the light travels a longer distnace. Therefore, they must measure a greater time interval. In other words, the Earth observer finds that the astronaut's clock runs slowly. If the speed of the spacecraft is v, then the distance that the light travels is cΔt=2D2+(vΔt2)2. Which means:

(263)Δt=2Dc2v2

Combining the two equations, we get the formula for time dilation:

(264)Δt=Δt01v2c2

Δt0 is the proper time interval, which is the time interval for the person that is at rest relative to the clock.

Simultaneity

Because of time dilation, events that are simulaneous in one frame of reference may not be simultaneous in another.

Length Contraction

Because time is different depeding on your frame of reference, since the speed remains constant, distance is therefore also different. Based off of v=xt, we can get:

(265)ΔL=ΔL01v2c2

Length contraction only occurs in the direction of motion.

ΔL0 is the proper length which is the length measured by what is NOT moving at relativistic speeds.

Relativistic Momentum

Apart from time and distances changing, momentum and energy also changes at exteremely high speeds.

(266)p=mv1v2c2

Relativistic momentum is always larger than nonrelativistic momentum.

Relativistic Addition of Velocities

Think back to how the traditional addition of velocities worked. At low speeds, this works. However, at high speeds, the addition of velocities changes also. This is because everything must stay relative to and relatively below light speed.

(267)vab=vac+vab1+vacvabc2

Relativistic Energy and the Equivalence of Mass and Energy

One of Einstein's principles is the mass-energy equivalence principle. This means that mass and energy are interchangable. The rest energy of something with mass is:

(268)E=mc2

For an object that is moving, its total energy is:

(269)E=mc21v2c2

Using theses two values, their difference is the kinetic energy of an object at high speeds:

(270)KE=mc21v2c2mc2

Using this formula, we can see that as v approaches c, KE approaches . Therefore, to go at the speed of light, you need an engine that is able to do infinite work, which is not possible.

Energy and Relativisic Momentum

The relationship between total relativistic energy and momentum is:

(271)E2=p2c2+m2c4

When a particle is at rest, p=0, so E=mc2.

For particles that have no mass, such as photons, the equation becomes:

(272)E=pc=hf=hcλ

27: Quantum Physics

Planck's Hypothesis

In 1900, Planck developed a formula for blackbody radiation, a perfect emitter that has no energy losses. He proposed that blackbody radiation was produced by submicroscopic charged oscillators, called resonators.

The resonators were allowed to only have certain discrete (quantized) energy levels, En.

(273)En=nhf
(274)h=6.626×1034Js

The Photoelectric Effect and the Particle Theory of Light

The photoelectric effect entails that light incident on certain metallic surfaces cause the emission of electrons from the surfaces. The emitted electrons are called photoelectrons.

Photoelectric Apparatus:

When the tube is dark, the ammeter reads 0A​. When the emitter is illuminated by light, the electrons get bumped off the emitter to the collector, completing the circuit, and causing a current to be measured. The current increases as the intensity of the light increases.

A successful explanation of the photoelectric effect was given by Einstein:

Experimentally, a linear relationship is observed between f and KEmax.

The cutoff wavelength is the minimum wavelength of light required to eject an electron. Referring to the previous equation, KEmax would therefore be 0, so the cutoff wavelength, λc, can be represented by:

(277)hcλc=W0

Applications:

A very useful application of this effect are photocells (solar panels). Think about how your scientific calculator works.

X-Rays

Roentgen discovered x-rays in 1895. X-rays are a type of light, so they travel at the speed of light, and they are not charged particles.

In 1912, Max von Laue suggested that, if x-rays really were a form of light with really small wavelengths, it should be possible to diffract them by using the regular atomic spacings of a crystal lattice as a diffraction granting.

Production of X-Rays X-rays are produced when high speed electrons are suddenly slowed down (like when a metal target is struck by electrons that have been accelerated through several thousand volts).

Production of X-Rays

Below is the graph of X-Ray Intensity per Unit Wavelength vs. Wavelength:

X-Ray Intensity per Unit Wavelength vs. Wavelength

Notice the peaks. These are called characteristic x-rays because they are characteristic of the target material.

As electrons at high speeds approach a surface, they decelerate due to positively charged nuclei in the material. This is represented by the Bremsstrahlung curve on the graph. Sometimes, the electron has the perfect conditions to eject another electron from the K shell of the surface material. This causes electrons in higher energy shells to drop down and fill in the gap, and, when an electron jumps down a gap, light, in this case, x-rays, is produced. This is represented by Kα and Kβ​ on the graph.

The electron that decelerated lost kinetic energy, which is now carried by the x-ray photon. In a very extreme and unrealistic case where all of an electron's kinetic energy is lost, the energy carried by the x-ray photon would be:

(278)E=eΔV=hfmax=hcλmin

eΔV is the energy carried by the electron, and the expression comes from ΔPE=qΔV.

Diffraction of X-Rays by Crystals

In theory, the wavelength of an electromagnetic wave can be measured if a grating having a suitable line spacing can be found. The spacing between the lines must be approximately equal to the wavelength.

The wavelength of x-rays happens to approximately equal the spacing between atoms. The structure of the atoms is determined by analyzing the positions and intensities of the various spots in the pattern.

X-Ray Diffraction

In a crystal structure, an x-ray that's deflected on a lower layer of the material travels a farther distance than one that is deflected on a higher layer.

Distance Traveled by X-Rays

The reflected beams will combine and produce constructive interference when the path difference is an integer multiple of the wavelength. This can be seen by Bragg's Law:

(279)2dsinθ=mλ

m​ is an integer multiplier. This formula can be used to calculate the spacing between atomic planes.

The Compton Effect

In 1923, Arthur Compton found out that, when x-rays were directed at a block of graphite, the scatter waves had lower energy. The amount of energy lost depended on the angle at which the rays scattered. This is called the Compton shift.

Compton Shift - The change in wavelength Δλ between a scatter x-ray and an incident x-ray.

Compton thought that an x-ray photon carries both energy and momentum. As the x-ray photon comes into contact with a specific material, some of its energy and momentum is transferred into an electron in the material.

(280)pincident photon=pscattered photon+precoiled electron

In chapter 26, we showed the following:

(281)p=mv1v2c2E=mc21v2c2

By manipulating these two equations, we can get the following equation for momentum:

(282)p=Evc2

In the case of lightwaves, however, v=c, so:

(283)p=Ec

Also remember that E=hf and c=fλ. Plugging these into the previous equation and simplifying, we get:

(284)p=hλ

The shift in wavelength of a photon is given by:

(285)Δλ=λfλi=hmelectronc(1cosθ)

θ is the angle of scatter.

The Dual Nature of Light and Matter

The Photoelectric Effect and Compton Effect offer evidence that light is a particle. However, since light exhibits interference and diffraction, it also acts as a wave.

Because of this, light is said to have a dual nature, having both wave and particle characteristics at once.

In 1924, de Broglie thought that since light waves exhibit particle characteristic, perhaps all forms of matter have a dual nature. This lead to the fact that electrons have a dual nature, too.

De Broglie's Hypothesis:

(286)λ=hp=hmv

The dual nature of matter is quite apparent, as the equation contains both particle concepts and wave concepts.

The Davisson-Germer Experiment

This experiment accidentally proved that electrons behave as waves.

The Uncertainty Principle

The Heisenberg Uncertainty Principle - If the measurement of the position of a particle is made with precision Δx, and a simultaneous measurement of the linear momentum of the particle is made with Δpx, the product of the two uncertainties can never be smaller than h4π.

(287)ΔxΔpx>h4π

All this pretty much says is that it is not possible to exactly precisely measure the position and momentum of a particle simultaneously.

Another form of this equation sets a limit on the accuracy of the measurement of energy within a time interval Δt.

(288)ΔEΔt>h4π

28: Atomic Physics

Early Models of the Atom

Atomic Spectra

Emission Spectra An evacuated glass tube filled with a gas with low pressure with will radiate light when a voltage is applied across it. When the light is analyzed with a spectrometer, an emission spectrum can be observed. Each element's emission spectrum is unique.

In 1885, when Balmer observed the emission spectrum of hydrogen, he found that the wavelengths of the lines can be described using the following equation, where n is an integer value of 3 or higher.

(289)1λ=RH(1221n2)

He didn't know it, but the 2 in 22 corresponded to the energy level that the electrons landed on, and n was the energy level that they started from. The correct base energy level is 1, but the jump is big enough that the emitted light is not on the visible light spectrum, so Balmer did not include this in his equation based off observation.

RH is the Rydberg constant.

(290)RH=1.0973732×107m1=191nm

Absorption Spectra

Apart from emitting light, an element can also absorb light at specific wavelengths, resulting in it having an absorption spectrum. The wavelengths of absorption spectra are the same as the wavelengths of emission spectra.

The Bohr Theory of Hydrogen

In 1913, Niels Bohr came up with a theory that explains why certain elements produce certain emission spectra. His theory applies to the hydrogen atom:

  1. The electron moves in circular orbits around the proton under the influence of the Coulomb force of attraction.

  2. Only certain orbits are stable.

  3. Radiation is emitted when the electron jumps from a high-energy state to a ground state.

    (291)ΔE=hf

    ΔE is the energy difference between the two energy levels.

  4. The size of allowed electron orbits is determined by a condition imposed on the electron's orbital angular momentum. The allowed orbits are where the angular momentum of the electron is an integer multiple of .

    (292)mevr=n=h2π

    This formula comes from the fact that Bohr said electrons travel in orbits. While in orbit, though, they have a wavelength. The position of the electron at the beginning of an orbit must match up with its position at the end of the orbit.

    Electron Wave Orbits

    The circumference of a circle with radius r is the path the electron takes, and this must equal an integer multiple of wavelengths:

    (293)2πr=nλ

    Remember the de Broglie wavelength: λ=hmev. Plugging this in, we end up with mevr=nh2π​.

The electric potential of the atom is represented by:

(294)PE=ke(e)(Ze)r

Z​ is the number of protons.

Simplifying, we get:

(295)PEatom=keZe2r

Assuming that the nucleus is at rest, the the total energy of the atoms is E=KEelectron+PEatom.

(296)E=12mev2keZe2r

Looking at just the electron, the attractive force is the centripetal force:

(297)keZe2r2=mv2r12mv2=KE=keZe22r

Plugging this value for kinetic energy back into the previous equation for energy, we get:

(298)E=keZe22r

Combining more of the equations in very arbitrary ways that are not worthy of a derivation, we end up with the formula for the radius of orbit of an electron:

(299)r=h2mekee2(n2Z)

This is basically saying that electrons can only exist in certain allowed orbits determined by the integer n.

Plugging this r equation into the energy equation, combining all the constants and simplifying, we get:

(300)E=(13.6eV)Z2n2

This is the energy corresponding to an energy level.

Ionization Energy Ionization energy is the energy required to completely remove an electron from the ground state.

Plugging this equation for energy into ΔE=hf and simplifying, we get:

(301)1λ=RHZ2(1nf21ni2)

Emission Series

Depending on which level electrons jump to and from, a different type of light is emitted. Each different type has a name.

Bohr's Correspondence Principle

Quantum mechanics is in agreement with classical physics as long as energy differences between quantized levels are very small. Basically, physics is weird inside an atom.

29: Nuclear Physics

Some Properties of Nuclei

The notation for nuclei, where X is the element symbol, is:

(302)ZAX

For example, the notation for helium with 2 neutrons and 2 protons is 24He.

Isotopes Atoms with the same element but a different number of neutrons.

The atomic mass of an element is the weighted average of all the naturally occurring isotopes of an element.

Charge and Mass

Relative Mass

(306)Relative mass of electrons to protons to neutrons1:1836:1839

Because the rest energy of a particle is given by E=mc2, it is often convenient to express a particle's mass in terms of its energy equivalent.

(307)1amu=931.5c2MeV

Nuclear Stability

A nucleus consisting of positively charge particles doesn't fall apart because of the nuclear force. This is an attractive, short-range force that acts between all nuclear particles. The protons attract each other through the nuclear force and repel each other through the Coulomb force.

The nuclear forces between proton-proton, proton-neutron, and neutron-neutron interactions are approximately the same.

As an atom gets more and more protons, it requires more neutrons to hold it together in order to overpower the Coulomb force. Because of this, as the elements get heavier on the periodic table, the more neutrons they have compared to protons.

Binding Energy

The total mass of a nucleus is always less than the combined mass of the particles that make it up (nucleons). This difference in mass Δm is known as the mass defect of the nucleus.

Binding Energy Diagram

E=mc2 shows that energy is related to mass. Therefore, when energy is put in the break apart a nucleus, the total mass of the separated nucleons is greater than that of the nucleus. The energy required to break a nucleus is called the binding energy.

(308)Ebinding=(Δm)c2

Radioactivity

In 1896, Becquerel accidentally discovered the radioactivity of uranium.

Radioactivity is the spontaneous emission of radiation from a nucleus.

Experiments that scientists of that time conducted suggested that radioactivity is the result of the decay (disintegration) of unstable nuclei. The rules of radioactivity are governed by the mass-energy equivalence: particles decay only when their combined products have less mass after decay than before (energy comes out).

Rutherford Rutherford showed that there are 3 types of radiation: alpha (α), beta (β), and gamma (γ​). He and his students discovered that the nucleus of an atom can be considered a point charge and mass and that most of the atomic mass is contained in the nucleus.

The Three Types of Radiation

  1. Alpha Particles (α) An alpha particle is a helium nuclei: 2 neutrons and 2 protons. Therefore, if a nucleus emits an alpha particle, it loses 2 neutrons and 2 protons.

    (309)ZAXZ2A4Y+24He

    Alpha particles are relatively weak. They can barely penetrate a sheet of paper. Even when traveling through air, it can only travel a few centimeters, as it grabs onto stray electrons and becomes a harmless helium atom.

  2. Beta Particles (β​) A beta particle is an electron or a positron (a positron is the antiparticle of an electron). There are two types of beta decay: beta-minus decay (electron is removed from nucleus) or beta-plus decay (positron is removed from nucleus). Beta-minus decay:

    (310)ZAXZ+1AY+10e

    Beta-plus decay:

    (311)ZAXZ1AY+10e

    Beta particles are a little stronger. They can penetrate a few millimeters of aluminum and eventually lose their energy through collisions.

  3. Gamma Particles (γ) Gamma particles are high energy photons (light). They are the strongest out of the three and can penetrate a few centimeters of lead.

The three types of radiation can be observed by putting the radioactive sample in a magnetic field:

Radiation in a Magnetic Field

This is a stability curve:

Stability Curve

The green line represents the point at which N=Z, and the red dots represent stable nuclei. Each labeled point on the graph represents and describes what type of radiation an element in that spot would emit. Each element's goal is to become stable.

The Decay Constant and Half Life

The rates of radioactive decay appear to be absolutely constant, even under extreme circumstances. Observation has shown that if a radioactive sample contains N nuclei, then the number of nuclei (ΔN) that decay in a time interval (Δt) is proportional to N.

(312)ΔNΔtN

Introducing the decay constant λ, the decay rate or activity R can be written as follows:

(313)R=|ΔNΔt|=λN

Alternatively, this can be viewed as a exponential differential equation:

(314)dNdt=λN

Isotopes with a large λ​ value decay rapidly, and vice-versa.

A unit of activity is a curie (Ci​).

(315)1Ci=3.7×1010decayss

The SI unit of activity is a becquerel (Bq).

(316)1Bq=1decays

The time it takes for a sample to decay to half of its original mass is called its half life.

Solving the above mentioned differential equation with the initial condition N(0)=N0, we get:

(317)N=N0eλt

Turning this into an exponential decay function by a factor of 12:

(318)N=N0(12)n=N0(12)tT

n represents the number of half lives that have passed. The half life can be obtained using:

(319)T=ln2λ

The Decay Process

Alpha Decay

If a nucleus emits an alpha particle, it loses 2 protons and 2 neutrons. The original nucleus is called the parent nucleus, and the nucleus after decay is called the daughter nucleus.

When one element changes into another, it is called spontaneous decay or transmutation.

The parent must contain more mass than the alpha particle and daughter combined as the excess mass is transformed into kinetic energy.

Beta Decay

When a nucleus undergoes beta decay, the daughter nucleus has the same number of nucleons, but the atomic number is changes by 1.

The Neutrino When an atom undergoes radioactive decay, the kinetic energy due to change in mass is actually less than it should be. This energy goes to a particle that we don't really see, called a neutrino. The neutrino (ν) has no charge, a mass smaller than that of an electron, a spin of 12, and a very weak interaction with matter (making it hard to detect). Additionally, ν represents an antineutrino.

Gamma Decay

The nucleus, like electrons, only exists in quantized energy states. When a nucleus undergoes radioactive decay, it is left in an excited state. When the nucleus transitions from this high energy state to a lower energy state, a high energy photon (gamma ray) is emitted.

Gamma decay does not cause transmutation, and there's really no way to predict it.

Top Problems

The top problems section is an assortment of the most challenging notes problems, homework problems, lab problems, FRQs, and some external problems I found on the internet to help review for each chapter.

If you are able to solve all the problems without issue, you are probably substantially prepared for the test.

I tried to include a variety of problems that hit all topics covered in each chapter.

Some questions may by slightly edited to fit the format of this document.

Chapter 2/3

  1. CJ Eq. of Kinem. for Const. Acc. Q6 A speed ramp at an airport is basically a large conveyor belt on which you can stand and be moved along. The belt of one ramp moves at a constant speed such that a person who stands still on it leaves the ramp 64s after getting on. Clifford is in a real hurry, however, and skips the speed ramp. Starting from rest with an acceleration of 0.37ms2, he covers the same distance as the ramp does, but in one-fourth the time. What is the speed at which the belt of the ramp is moving? (0.74ms)

  2. CJ Eq. of Kinem. for Const. Acc. Q9 A locomotive is accelerating at 1.6ms2. It passes through a 20.0m wide crossing in a time of 2.4s. After the locomotive leaves the crossing, how much time is required until its speed reaches 32ms? (14s)

  3. CH 2 WS #4 Q11 A peregrine falcon dives at a pigeon. The falcon starts downward from rest and falls with free-all acceleration. If the pigeon is 76.0m below the initial position of the falcon, how long does it take the falcon to reach the pigeon? Assume that the pigeon remains at rest. (3.94s)

  4. CJ Proj. Mot. Q7 The drawing shows two planes each about to drop an empty fuel tank. At the moment of release each plane has the same speed of 135ms, and each tank is at the same height of 2.00km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0 above the horizontal and the other is flying at an angle of 15.0 below the horizontal. Find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is from (a) plane A (239ms@57.1) (b) plane B (239ms@57.1) In each part, give the directional angles with respect to the horizontal. CJ Proj. Mot. Q7

  5. CJ Proj. Mot. Q11 A rocket is fired at a speed of 75.0ms from ground level, at an angle of 60.0 above the horizontal. The rocket is fired toward an 11.0m high wall, which is located 27.0m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall? (33.2m)

  6. CH 3 WS #3 Q8 Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One paddles directly upstream (and moves upstream), whereas the other paddles directly downstream. If downstream is the positive direction, an observer on shore determines the velocities of the two canoes to be 1.2ms and +2.9ms, respectively. (a) What is the speed of the water relative to shore? (0.85ms) (b) What is the velocity of each canoe relative to the water? (2.05ms, 2.05ms)

  7. CH 2&3 Review Sup. Prob. Q19 A river has a steady speed of 0.5ms. A student swims upstream a distance of 1.0km and returns to the starting point. If the student can swim at a speed of 1.2ms in still water, how long does the trip take? Compare this with the time the trip would take if the water were still. (2017s, 1667s)

Chapter 4

  1. CH 4 WS #1 Q14 A 5.0g bullet leaves the muzzle of a rifle with the speed of 320ms. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.82m long barrel of the rifle? (310N)

  2. CH 4 WS #3 Q7 A 150N bird feeder is supported by three cables as shown in the figure. Find the tension of each cable. (75N, 130N) CH 4 WS #3 Q7

  3. CH 4 WS #5 Q8 Two blocks of masses 40.0kg and 20.0kg are stacked on a table with the heavier block on top. The coefficient of static friction is 0.600 between the two blocks and 0.300 between the bottom block and the table. A horizontal force is slowly applied to the top block until one of the blocks moves. (a) Where does slippage occur first: between the two blocks or between the bottom block and the table? Explain. (bottom block) (b) What value of the coefficient of static friction between the bottom block and the table would change the answer to part (a)? (0.4)

  4. CH 4 Test Review Q16 A toboggan slides down a hill and has a constant velocity. The angle of the hill is 8.00 with respect to the horizontal. What is the coefficient of kinetic friction between the surface of the hill and the toboggan? (0.141)

Chapter 5

  1. CJ Cons. of E. Q14 Two pole-vaulters just clear the bar at the same height. The first lands at a speed of 8.90ms, and the second lands at a speed of 9.00ms. The first vaulter clears the bar at a speed of 1.00ms. Ignore air resistance and friction and determine the speed at which the second vaulter clears the bar. (1.7ms)

  2. CH 5 WS #4 Q9 A surprising demonstration involves dropping an egg from a third-floor window so that the egg lands on a foam-rubber pad without breaking. If a 56.0g egg falls (from rest) 12.0m, and the 5.00cm thick foam pad stops it in 6.25ms, by how much is the pad compressed? (Note: Assume constant upward acceleration as the egg compresses the foam-rubber pad.) (4.79cm)

  3. CH 5 WS #5 Q7 A 5.00g feather is held 50.0cm above a table. It is dropped from rest and subjected to the non-conservative, constant force of air friction f. When the feather reaches the table, it is falling at a speed of 10.0cms. Calculate (a) the magnitude of the force of air resistance, f, and (0.049N) (b) the mechanical energy lost by the feather due to air friction. (2.448×102J​)

  4. CH 5 WS #6 Q7 While running, a person dissipates about 0.60J of mechanical energy per step per kilogram of body mass. If a 60kg person develops a power of 70W during a race, how fast is the person running? Assume that a running step is 1.5m long. (2.9ms​)

  5. CH 5 WS #6 Q11 A 650kg elevator starts from rest. It moves upward for 3s with constant acceleration until it reaches its cruising speed, 1.75ms. (a) What is the average power of the elevator motor during this period? (5910W) (b) How does this compare with its power during an upward cruise with constant speed? (11200W​)

  6. CH 5 WS #1 Q10 A 70.0kg base runner begins his slide into second base when moving at a speed of 4.0ms. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base. (a) How much mechanical energy is lost due to friction acting on the runner? (560J) (b) How far does he slide? (1.2m)

  7. CH 5 WS #2 Q14 A 10kg crate is pulled up a rough incline with an initial speed of 1.5ms. The pulling force is 100N parallel to the incline, which makes an angle of 20 with the horizontal. If the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 5.0m, (a) how much work is done by the gravitational force? (168J) (b) How much work is done by the 100-N force? (500J) (c) What is the change in kinetic energy of the crate? (148J) (d) What is the speed of the crate after it is pulled 5.0 m? (5.64ms​)

Chapter 6

  1. CH 6 WS #2 Q13 A 730N man stands in the middle of a frozen pond of radius 5.0m. He is unable to get to the other side because the lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2kg physics book horizontally toward the north shore, at the speed of 5.0ms. How long does it take him to reach the south shore? (62s​)

  2. CH 6 WS #4 Q5 A 2.0g particle moving at 8.0ms makes a perfectly elastic head-on collision with a resting 1.0g object. (a) Find the speed of each after the collision. (10.7ms, 2.67ms) (b) If the stationary particle has a mass of 10g, find the speed of each particle after the collision. (2.67ms, 5.33ms) (c) Find the final kinetic energy of the incident 2.0g particle in the situations described in (a) and (b). In which case does the incident particle lose more kinetic energy? (7.11×103J, 2.8×102J​, case a)

  3. CH 6 Supplemental Problems Q4 An estimated force-time curve for a baseball struck by a bat is shown in the figure. From this curve, determine (a) the impulse delivered to the ball and (12Ns) (b) the average force exerted on the ball. (2.4N) CH 6 Supplemental Problems Q4

Chapter 7

  1. CJ Chosen Tan. V&A Q15 A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a constant tangential speed of 1.20ms on its circular path. Find the linear speed with which the bucket moves down the well. (0.300ms) CJ Chosen Tan. V&A Q15

  2. CH 7 WS #2 Q4 An air puck of mass 0.25kg is tied to a string and allowed to revolve in a circle of radius 1.0m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of 1.0kg is tied to it (see figure). The suspended mass remains in equilibrium while the puck on the tabletop revolves. (a) What is the tension in the string? (9.8N) (b) What is the centripetal force acting on the puck? (9.8N) (c) What is the speed of the puck? (6.3ms) CH 7 WS #2 Q4

  3. CJ Chosen Cent. F, Banked Curves, G Q15 A stone has a mass of 6.0×103kg and is wedged into the tread of an automobile tire, as the drawing shows. The coefficient of static friction between the stone and each side of the tread channel is 0.90. When the tire surface is rotating at a maximum speed of 13ms, the stone flies out of the tread. The magnitude FN of the normal force that each side of the tread channel exerts on the stone is 1.8N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire. (0.31m) CJ Chosen Cent. F, Banked Curves, G Q15

  4. CJ Chosen Cent. F, Banked Curves, G Q17 On a banked race track, the smallest circular path on which cars can move has a radius of 112m, while the largest has a radius of 165m, as the drawing illustrates. The height of the outer wall is 18m. Find (a) the smallest and (19ms) (b) the largest speed at which cars can move on this track without relying on friction. (23ms) CJ Chosen Cent. F, Banked Curves, G Q17

  5. CJ Chosen Cent. F, Banked Curves, G Q21 Saturn has an equatorial radius of 6.00×107m and a mass of 5.67×1026kg. (a) Compute the acceleration of gravity at the equator of Saturn. (10.5ms2) (b) What is the ratio of a person’s weight on Saturn to that on earth? (1.07)

  6. CH 7 WS #3 Q15 A pail of water is rotated in a vertical circle of radius 1.00m (the approximate length of the person’s arm). What must be the minimum speed of the pail at the top of the circle if no water is to spill out? (3.13ms)

  7. CH 7 Notes P14 Q5 A projectile is fired straight up from the South Pole of Earth with an initial speed vi=8.0kms. Find the maximum height it reaches, neglecting air resistance. The mass of Earth is 5.98×1024kg and the radius of Earth is 6.37×106m. (6.66×106m)

  8. CH 7 Notes P15 Q6 How much energy is required to lift a 9000kg vehicle from Earth's surface to the height of the International Space Station, 400km above the surface? (3.35×1010J)

  9. CH 7 Notes P15 Q7 Compute the escape velocity (in ms) for Earth if its mass is 5.98×1024kg and its radius is 6.37×106m. (1.12×104ms)

  10. CH 7 Notes P17 Q1 A satellite of mass m moves in a circular orbit about the Earth with a constant speed of v and a height of h=1000km above the Earth's surface. Find the orbital speed of the satellite in mih. The radius of the Earth is 6.38×106m, and its mass is 5.98×1024kg. (16400mih)

  11. External If Saturn is, on average, 9 times farther from the Sun than Earth is, how long is its year in terms of Earth years? (Hint: use Kepler's Third Law.) (27yrs​)

  12. External An object starts from rest at point A and rolls along the frictionless track, though the loop as shown in the figure. What is the direction of the net force acting on the block at point C? (down and to the left) External

Chapter 8

  1. CH 8 WS #1 Q11 The person in the figure weighs 800N. The forces F1 and F2 have the magnitudes of 100N and 900N, respectively. Assume that the force of gravity acts downward through point A, as shown. Determine the net torque on the person about axes though the points A, B, and C perpendicular to the plane of the paper. (Hint: Replace F1 and F2 with their horizontal and vertical components.) (207Nm, 145Nm, 95.7Nm) CH 8 WS #1 Q11

  2. CJ Chosen Rot. Equil. Q11 The drawing shows a person (weight, W=584N) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position. (196N, 96N) CJ Chosen Rot. Equil. Q11

  3. CH 8 WS #3 Q8 A water molecule consists of an oxygen atom with two hydrogen atoms bound to it as shown in the figure. The bonds are 0.100nm in length, and the angle between the two bonds is 106. Use the coordinate axes shown, and determine the location of the center of gravity of the molecule. Take the mass of an oxygen atom to be 16 times the mass of a hydrogen atom. ((6.69×103m,0m)) CH 8 WS #3 Q8

  4. CH 8 WS #4 Q9 A 5.00kg cylindrical reel with a radius of 0.600m and a frictionless axle starts from rest and speeds up uniformly as a 3.00kg bucket falls into a well, making a light rope unwind from the reel (see figure). The bucket starts from rest and falls for 4.00s. (a) What is the linear acceleration of the falling bucket? (5.35ms2) (b) How far does it drop? (42.8m) (c) What is the angular acceleration of the reel? (8.91s2) CH 8 WS #4 Q9

  5. CH 8 WS #5 Q11 In the figure, the sliding block has a mass of 0.850kg, the counterweight has a mass of 0.420kg, and the pulley is a uniform solid cylinder with a mass of 0.350kg and an outer radius of 0.0300m. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of 0.820ms toward the pulley when it passes through a photogate. (a) Use energy methods to predict the speed of the block after it has moved to a second photogate 0.700 m away. (1.63ms) (b) Find the angular speed of the pulley at the same moment. (54.2s2) CH 8 WS #5 Q11

  6. CH 8 WS #6 Q10 A playground merry-go-round of radius 2.00m has a moment of inertia I=275kgm2 and is rotating about a frictionless vertical axle. As a child of mass 25.0kg stands at a distance of 1.00m from the axle, the system (merry-go-round and child) rotates at the rate of 14.0revmin. The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child reaches the edge? (1.17s1)

  7. CH 8 WS #6 Q15 A 60.0kg woman stands at the rim of a horizontal turntable having a moment of inertia of 500kgm2 and a radius of 2.00m. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50ms relative to the Earth. (a) In what direction and with what angular speed does the turntable rotate? (0.360s1) (b) How much work does the woman do to set herself and the turntable into motion? (99.9J)

  8. External A traffic light hangs from a pole as shown in the figure. The uniform aluminum pole AB is 7.20m long and has a mass of 12.0kg. The mass of the traffic light is 21.5kg. Determine (a) the tension in the horizontal massless cable CD (410N) (b) the magnitude of the force exerted by the pivot A on the pole (525N) External

  9. External If the mass of Mars is 0.107 times that of Earth, and its radius is 0.530 that of Earth, estimate the gravitational acceleration at the surface of Mars. (3.73ms2​)

  10. AP Mechanics Review 2 Q9c A box of uniform density weighing 125N moves in a straight line with constant speed along a horizontal surface. The coefficient of sliding friction is 0.65. A horizontal force F, applied at a height 53m above the surface as shown in the diagram, is just sufficient to cause the box to begin to tip forward about an axis through point P. The box is 1m wide and 2m high. Calculate the force F. (37.5N​) AP Mechanics Review 2 Q9c

  11. 2022 Practice Exam 3 MCQ Q9 Two objects, X and Y, experience external net torques that vary over a period of 5 seconds. Object X has a moment of inertia I0, and Object Y has a moment of inertia 2I0. The average value of the magnitude of the external net torque exerted on Object X from time t=0 to t=5 is τX. Similarly, the average value for Object Y is τY. The magnitudes of the angular momenta L of Objects X and Y versus t are shown in the graph. Which of the following expressions correctly relates τY to τX​? (b) 2022 Practice Exam 3 MCQ Q9 a. τY=4τX b. τY=2τX c. 2τY=τX d. 4τY=τX

Chapter 9

  1. CH 9 Notes P13 Q1 A solid, square, pinewood raft measures 4.0m on a side and is 0.30m thick. (Density of pine is 550kgm3) a) Determine whether the raft floats in water. (yes) b) If so, how much of the raft is beneath the surface? (0.17m)

  2. CH 9 Notes P17 Q2 In the condition known as atherosclerosis, a deposit or atheroma forms on the arterial wall and reduces the opening through which blood can flow. In the carotid artery in the neck, blood flows three times faster through a partially blocked region than it does through an unobstructed region. Determine the ratio of the effective radii of the artery at the two places. (1.7)

  3. CH 9 Notes P20 Q1 A nearsighted sheriff fires at a cattle rustler with his trusty six-shooter. Fortunately for the cattle rustler, the bullet misses him and penetrates the town water tank to cause a leak. If the top of the tank is open to the atmosphere, determine the speed at which the water leaves the hole when the water level is 0.500m above the hole. (3.13ms)

  4. CH 9 Notes P22 Q1 An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Suppose that, because of an aneurysm, the cross sectional area A1 of the aorta increases to a value A2=1.7A1. The speed of the blood (ρ=1060kgm3) through a normal portion of the aorta is v1=0.40ms. Assuming the aorta is horizontal (the person is lying down), determine the amount by which the pressure in the enlarged region exceeds that in the normal region. (54Pa)

  5. CH 9 WS #3 Q8 An empty rubber balloon has a mass of 0.0120kg. The balloon is filled with helium at a density of 0.181kgm3. At this density the balloon has a radius of 0.500m. If the filled balloon is fastened to a vertical line, what is the tension in the line? (ρair=1.29kgm3) (5.57N)

  6. CH 9 WS #4 Q9 A fountain sends a stream of water 20m up into the air. If the base of the stream is 10cm in diameter, what power is required to send the water to this height? (30kW)

  7. CH 9 WS #4 Q12 A hypodermic syringe contains a medicine with the density of water (see figure). The barrel of the syringe has a cross-sectional area of 2.50×105m2. The cross-sectional area of the needle is 1.00×108m2. In the absence of a force on the plunger, the pressure everywhere is 1.00atm. A force, F, of magnitude 2.00N is exerted on the plunger, making medicine squirt from the needle. Determine the medicine’s flow speed. Assume that the pressure in the needle remains equal to 1.00atm, that the syringe is horizontal, and that the speed of the emerging fluid is the same as the speed of the fluid in the needle. (assume v1 (speed of fluid in the syringe) =0 when compared to v2 (speed of fluid in the needle) due to the difference in their cross- sectional areas) (12.6ms) CH 9 WS #4 Q12

  8. External When a golf ball with a radius of 1.5cm is hung from a spring scale in air, the spring scale has a reading of 50g. A beaker is filled with water and placed on a balance, weighing 500g in total. While still hanging from the spring scale, the golf ball is fully submerged in the water. Find the reading on the spring scale and on the balance. (36g, 514g)

Chapter 10

  1. CH 10 Notes P9 Q3 A sealed glass bottle at 27C contains air at atmospheric pressure and has a volume of 30cm3. The bottle is then tossed into an open fire. When the temperature of the air in the bottle reaches 200C, what is the pressure inside the bottle? Assume that any volume changes of the bottle are small enough to be negligible. (1.6atm)

  2. CH 10 Notes P9 Q5 Verify that one mole of oxygen occupies a volume of 22.4L at 1atm and 0C.

  3. CH 10 Notes P14 Q1 Air is primarily a mixture of nitrogen, N2, and oxygen, O2. Assume that each behaves as an ideal gas. At a temperature of 293K, find the average translational kinetic energy for each type of molecule. (6.07×1021J)

  4. CH 10 Notes P14 Q2 In the previous example, we found that the average translational kinetic energy of each molecule in air is 6.07×1021J at 293K. Determine the rms speed of the nitrogen (molecular mass =28.0amu) and oxygen (molecular mass =32.0amu) molecules in the air at this temperature. (511ms, 478ms)

  5. CH 10 Notes P8 Q1 In scuba diving, a greater water pressure acts on a diver at greater depths. The air pressure inside the body cavities (e.g. lungs, sinuses) must be maintained at the same pressure as that of the surrounding water, otherwise they might collapse. A special valve automatically adjusts the pressure of the air breathed from a scuba combustion tank to ensure that the air pressure equals the water pressure at all times. There is some scuba gear that consists of a 0.0150m3 tank filled with compressed air at an absolute pressure of 2.02×107Pa. Assuming that air is consumed at a rate of 0.0300m3 per minute and that the temperature is the same at all depths, determine how long the diver can stay under at a depth of a) 10.0m and (49.5min) b) 30.0m. (24.6min)

  6. CH 10 WS #3 Q7 The total random translational kinetic energy of the water molecules in a 20L container is 4.8×106J; the total random translational kinetic energy of the water molecules in a 360000L swimming pool is 7.3×1010J. Which body of water, the container or the pool, is at the higher temperature? (container)

  7. External A pressure of 1.0×107mmHg is achieved in a vacuum system. How many gas molecules are present per liter volume if the temperature is 293K? (760mmHg=1atm, R=0.082LatmmolK and NA=6.02×1023mol1) (3.3×1012L1)

Chapter 11

  1. CH 11 WS #1 Q10 A 200g block of copper (cCu=387JkgC) at a temperature of 90C is dropped into 400g of water at 27C. The water is contained in a 300g glass (cglass=837JkgC) container. What is the final temperature of the mixture? (30C)

  2. CH 11 WS #2 Q2 A large block of ice at 0C has a hole chipped in it, and 400g of aluminum (cAl=900JkgC) pellets at a temperature of 30C are poured into the hole. How much of the ice melts? (32g)

  3. CH 11 WS #2 Q4 What mass of steam that is initially at 120C is needed to warm 350g of water in a 300g aluminum (cAl=900JkgC) container from 20C to 50C? (21g)

Chapter 12

  1. CH 12 Notes P3 Q1 A gas is contained in a cylinder with a moveable piston on top. The gas is at a pressure of 8000Pa and the piston has an area of 0.10m2. As heat is slowly added to the gas, the piston is pushed up a distance of 4.0cm. Calculate the work done on the surroundings by the expanding gas. Assume that the pressure remains constant. (32J)

  2. CH 12 Notes P4 Q4 Gas in a cylinder moves a piston with area 0.20m2 as energy is slowly added to the system. If 2.00×103J of work is done on the environment and the pressure of the gas in the cylinder remains constant at 1.01×105Pa, find the displacement of the piston. (9.90×102m)

  3. CH 12 Notes P5 Q1 A system absorbs 1500J of heat from its surroundings and performs 2200J of work on its surroundings. Determine the change in the internal energy of the system. (700J)

  4. CH 12 Notes P6 Q3 The temperature of three moles of a monatomic ideal gas is reduced from Ti=540K to Tf=350K by two different methods. In the first method, 5500J of heat flows into the gas, while in the second method, 1500J of heat flows into it. In each case, find a) the change in the internal energy of the gas and (7100J) b) the work done by the gas. (12600J, 8600J)

  5. CH 12 Notes P11 Q2 Find the efficiency of an engine that introduces 2000J of heat during the combustion phase and loses 1500J at exhaust. (25%)

  6. CH 12 Notes P14 Q1 Water near the surface of a tropical ocean has a temperature of 298.2K, while water 700m beneath the surface has a temperature of 280.2K. It has been proposed that the warm water be used as a hot reservoir and the cool water as the cold reservoir of a heat engine. a) Find the maximum possible efficiency for such an engine. (6.0%) b) Determine the minimum input heat, QH, that would be needed if a number of these engines were to produce an amount of work equal to the 8.1×1019J of energy that the U.S. consumed in 1978. (1.4×1021J)

  7. CH 12 Notes P15 Q4 The highest theoretical efficiency of a gasoline engine, based on the Carnot cycle, is 30%. If this engine expels its gases into the atmosphere, which has a temperature of 300K, what is the temperature in the cylinder immediately after combustion? (430K)

  8. CJ Chosen 1st Law Prob. Q8 Three moles of an ideal monatomic gas are at a temperature of 345K. Then, 2438J of heat is added to the gas, and 962J of work is done on it. What is the final temperature of the gas? (436K)

  9. CH 12 Notes P19 Q1 1200J of heat flows spontaneously from a hot reservoir at 650K to a cold reservoir at 350K. Determine the amount by which this irreversible process changes the entropy of the universe, assuming that no other changes occur. (1.6JK)

  10. CH 12 WS #3 Q5 Two 2000kg cars, both traveling at 20ms, undergo a head-on collision and stick together. Find the entropy change of the Universe during the collision if the temperature is 23C. (2700JK)

Chapter 13

  1. CH 13 Notes P8 Q1 A 1.3×103kg car is constructed on a frame supported by four springs. Each spring has a spring constant of 2.00×104Nm. If two people riding in the car have a combined mass of 160kg, find the frequency of vibration of the car when it is driven over a pothole in the road. Find also the period and angular frequency. Assume the weight is evenly distributed. (1.18Hz, 0.847s, 7.41s1​)

  2. CH 13 Notes P15 Q1 AM and FM radio waves are transverse waves consisting of electric and magnetic disturbances traveling at a speed of 3.00×108ms. A station broadcasts an AM radio wave whose frequency is 1230×103Hz (1230kHz on the dial) and an FM radio wave whose frequency is 91.9×106Hz (91.9MHz on the dial). Find the distance between adjacent crests in each wave. (244m, 3.26m​)

  3. CH 13 Notes P17 Q1 A uniform string has a mass M of 0.0300kg and a length L of 6.00m. Tension is maintained in the string by suspending a block of mass m=2.00kg from one end. (a) Find the speed of the transverse wave pulse on this string. (62.6ms) (b) Find the time it takes the pulse to travel from the wall to the pulley. (0.0799s) CH 13 Notes P17 Q1

  4. CH 13 Notes P17 Q2 To what tension must a string with mass 0.0100kg and a length 2.50m be tightened so that waves will travel on it at a speed of 125ms? (62.5N​)

  5. CH 13 WS #3 Q14 A light string mass 10.0g and length L=3.00m has its ends tied to two walls that are separated by the distance D=2.00m. Two masses, each of mass m=2.00kg, are suspended from the string as in the figure. If a wave pulse is sent from point A, how long does it take to travel to point B? (0.0329s) CH 13 WS #3 Q14

Chapter 15

  1. CH 15 Notes P3 Q1 The electron and proton of a hydrogen atom are separated (on average) by a distance of about 5.3×1011m. Find the magnitudes of the electric force and gravitational force that each particle exerts on the other, and the ratio of the electric force Fe to the gravitational force Fg. (8.2×108N, 3.6×1047N, 2.27×1039)

  2. CH 15 Notes P4 Q3 Three charges lie along the x-axis as in the figure. The positive charge q1=15µC is at x=2.0m, and the positive charge q2=6.0µC is at the origin. Where must a negative charge q3 be placed on the x-axis so that the resultant electric force on it is zero? (0.77m away from q2​)

  3. CH 15 WS #1 Q12 A typical lead-acid storage battery contains sulfuric acid, HA2SOA4, which breaks down into 2H+SOA4 and each molecule delivers two electrons to the external circuit. If the battery delivers a total charge of 2.0×105C, how many grams of sulfuric acid are used up? (102g​)

  4. CH 15 Notes P6 Q8 In the Bohr model of the hydrogen atom, the electron is in a circular orbit about the nuclear proton at a radius of 5.29×1011m. The mass of an electron is 9.11×1031kg. Determine the speed of the electron. (2.19×106ms​)

  5. CH 15 Notes P8 Q1 Tiny droplets of oil acquire a small negative charge while dropping through a vacuum (pressure =0) in an experiment. An electric field of magnitude 5.92×104NC​ points straight down. (a) One particular droplet is observed to remain suspended against gravity. If the mass of the droplet is 2.93×1015kg, find the charge carried by the droplet. (4.85×1019C) (b) Another droplet of the same mass falls 10.3cm from rest in 0.250s, again moving through a vacuum. Find the charge carried by the droplet. (3.22×1019C)

  6. CH 15 WS #4 Q11 Each of the protons in a particle beam has a kinetic energy of 3.25×1015J. What are the magnitude and direction of the electric field that will stop these protons in a distance of 1.25m? (1.63×104NC)

Chapter 16

  1. CH 16 Notes P4 Q6 In atom smashers (also known as cyclotrons and linear accelerators), charged particles are accelerated in much the same way they are accelerated in TV tubes: through potential differences. Suppose a proton (mass =1.67×1027kg) is injected at a speed of 1.00×106ms between two plates 5.00cm apart. The proton subsequently accelerates across the gap and exits through the opening. (a) What must the electric potential difference be if the exit speed is to be 3.00×106ms? (41800V) (b) What is the magnitude of the electric field between the plates? (8.36×105NC)

  2. CH 16 Notes P5 Q7 Suppose electrons (mass =9.11×1031kg) in a TV tube are accelerated through a potential difference of 2.00×104V from the heated cathode (negative electrode), where they are produced, toward the screen, which also serves as the anode (positive electrode), 25.0cm away. (a) At what speed would the electrons impact the phosphors on the screen? Assume they accelerate from rest, and ignore resistive effects. (8.38×107ms​) (b) What's the magnitude of the electric field, if it is assumed constant? (8.00×104NC​)

  3. CH 16 Notes P10 Q5 Three point charges are initially infinitely far apart. They are then brought together and placed at the corners of an equilateral triangle. Each side of the triangle has a length of 0.50m. Determine the electric potential energy of the triangular group. In other words, determine the amount by which the electric potential energy of the group differs from that of the three charges in the initial, infinitely separated locations. (0.14J) CH 16 Notes P10 Q5

  4. CH 16 Notes P16 Q4 (a) Calculate the equivalent capacitance between a and b for the combination of capacitors shown in the figure. All capacitances are in microfarads. (6.0µF) (b) If a 12V battery is connected across the system between points a and b, find the charge on the 4µF capacitor and the voltage drop across it. (24µC, 6.0V) (c) Find the charge on the 8µF capacitor and the voltage drop across it. (48µC, 6.0V) (d) Find the charge on the 6µF capacitor and the voltage drop across it. (36µC, 6.0V) CH 16 Notes P16 Q4

  5. CH 16 Notes P17 Q1 A fully charged capacitor defibrillator contains 1.20kJ of energy stored in a 1.10×104F capacitor. In a discharge through a patient, 6.00×102J of electrical energy are delivered in 2.50ms. (a) Find the voltage needed to store 1.20kJ in the unit. (4.67×103V) (b) What average power is delivered to the patient? (2.40×105W​)

  6. CH 16 WS #4 Q6 A parallel-plate capacitor has 2.00cm2 plates that are separated by 5.0mm with air between them. If a 12.0V battery is connected to this capacitor, how much energy does it store? (2.55×1011J)

Chapter 17

  1. CH 17 Notes P1 Q2 Suppose 6.40×1021 electrons pass through a wire in 2.00min. Find the current. (8.53A​​)

  2. CH 17 WS #1 Q11 A teapot with a surface area of 700cm2 is to be silver plated. It is attached to the negative electrode of an electrolytic cell containing silver nitrate (Ag+NO3). If the cell is powered by a 12.0V battery and has a resistance of 1.80, how long does it take to build up a 0.133mm layer of silver on the teapot? (Density of silver =10.5×103kgm3; Molar Mass of Ag =107.9gmol) (1.31×104s=3.64h​)

  3. CH 17 Notes P9 Q3 An electric heater is operated by applying a potential difference of 50.0V to a nichrome wire of total resistance 8.00. (a) Find the current carried by the wire and the power rating of the heater. (6.25A, 313W) (b) Using this heater, how long would it take to heat 2.5×103 moles of diatomic gas from a chilly 10.0C to 25.0C? Take the molar specific heat at constant volume of air to be 52R. (2.49×103s​)

Chapter 18

  1. CH 18 WS #2 Q7 Two 1.50V batteries – with their positive terminals in the same direction – are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.255, the other an internal resistance of 0.153. When the switch is closed, a current of 0.600A passes through the lamp. (a) What is the lamp’s resistance? (4.59) (b) What fraction of the power dissipated is dissipated in the batteries? (0.082​)

  2. CH 18 WS #2 Q8 An unmarked battery has an unknown internal resistance. If the battery is connected to a fresh 5.60V battery (negligible internal resistance) positive to positive and negative to negative, the current through the circuit is 10.0mA. If the polarity of the unknown battery is reversed, the current increases to 25.0mA. Determine the source voltage and internal resistance of the unknown battery. Assume that in each case the direction of the current is negative to positive in the 5.60V battery. (2.4V, 320​)

  3. CH 18 WS #2 Q12 Determine the potential difference, Vab, for the circuit in the figure. (5.4V) CH 18 WS #2 Q12

  4. CH 18 Notes P12 Q1 An uncharged capacitor and a resistor are connected in series to a battery. If ϵ=12.0V, C=5.00µF, and R=8.00×105, find (a) the time constant of the circuit, (4.00s) (b) the maximum charge on the capacitor, (60.0µC) (c) the charge on the capacitor after 6.00s, (4.66×105C) (d) the potential difference across the resistor after 6.00s, (2.68V) (e) and the current in the resistor at that time. (3.4×106A​​)

  5. AP Electricity Review 2 Q6e The circuit above contains a battery with negligible resistance, a closed switch S, and three resistors each with a resistance R or 2R. Use ϵ=18V and R=150. The switch S is opened, resistor RB is removed and replaced by a capacitor of capacitance 2.0×106F, and the switch S is again closed. Calculate the charge on the capacitor after all the currents have reached their final steady-state values. AP Electricity Review 2 Q6e

Chapter 19

  1. CH 19 WS #2 Q13 Consider the mass spectrometer shown schematically in the figure. The electric field between the plates of the velocity selector is 950Vm, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.93T. Calculate the radius of the path in the system for singly charged ions with mass m=2.18×1026kg. (1.50×104m) CH 19 WS #2 Q13

  2. CH 19 Notes P15 Q2 Two wires, each having a weight per unit length of 1.00×104Nm, are parallel with one directly above the other. Assume that the wires carry currents that are equal in magnitude and opposite in direction. The wires are 0.10m apart, and the sum of the magnetic force and gravitational force on the upper wire is zero. Find the current in the wires, neglecting Earth's magnetic field. (7.07A​)

  3. CH 19 WS #3 Q6 The figure is a cross-sectional view of a coaxial cable (such as a VCR cable). The center conductor surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another layer of rubber. The current in the inner conductor is 1.00A. out of the page, and the current in the outer conductor is 3.00A into the page. Using Ampere’s law, determine the magnitude and direction of the magnetic fields at points A and B. (2×104T CCW, 1.33×104T CW) CH 19 WS #3 Q6

Chapter 20

  1. CH 20 WS #1 Q9 A circular wire loop of radius 0.50m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.40T. If in 0.10s the wire is reshaped from a circle into a square, but remains in the same plane, what is the magnitude of the average induced emf in the wire during this time? (0.67V)

  2. CH 20 WS #1 Q15 A square, single-turn coil 0.20m on a side placed with its plane perpendicular to a constant magnetic field. An emf of 18mV is induced in the winding when the area of the coil decreases at a rate of 0.10m2s. What is the magnitude of the magnetic field? (0.18T​)

  3. CH 20 Notes P7 Q1 An airplane with a wingspan of 30.0m flies due north at a location where the downward component of the Earth's magnetic field is 0.600×104T. There is also a component pointing due north which has a magnitude of 0.470×104T. (a) Find the difference in potential between the wingtips when the speed of the plane is 2.50×102ms. (0.450V​) (b) Which wing tip is positive? (west wing tip)

  4. CH 20 Notes P8 Q3 The sliding bar in the following figure has a length of 0.500m and moves at 2.00ms in a magnetic field of magnitude 0.250T. (a) Using the concept of motional emf, find the induced voltage in the moving rod. (0.250V) (b) If the resistance in the circuit is 0.500, find the current in the circuit and the power delivered to the resistor. (0.500A, 0.125W) (c) Calculate the magnetic force on the bar. (6.25×102N in x direction) (d) Use the concepts of work and power to calculate the applied force. (6.25×102N) CH 20 Notes P8 Q3

  5. CH 20 WS #2 Q3 A circular coil, enclosing an area of 100cm2, is made of 200 turns of copper wire, as shown in the figure. Initially, a 1.1T uniform magnetic field points perpendicularly upward through the plane of the coil. The direction of the field then reverses so that the final magnetic field has a magnitude of 1.1T and points downward through the coil. During the period in which the field is changing direction, how much charge flows through the coil if the coil is connected to a 5.0 resistor as shown? (0.88C​) CH 20 WS #2 Q3

  6. 2017 International Practice Exam MCQ Q9 A rectangular loop of copper wire is attached to a cart by an insulating rod. The cart is moving at constant speed when it enters a region containing a uniform magnetic field that is perpendicular to the plane of the loop and directed into the page, as shown below. Frictional losses are negligible. Which of the following correctly describes the speed of the cart as it moves into, through, and out of the field? (d) 2017 International Practice Exam MCQ Q9 a. The speed remains constant. b. The speed continually decreases. c. The speed decreases as the cart enters the field and increases as it leaves the field but is constant while it is completely inside the field. d. The speed decreases as the cart enters and leaves the field but is constant while it is completely inside the field.

Chapter 21

  1. CH 20 WS #3/CH 21 WS #1 Q10 A transformer on a pole near a factory steps the voltage down from 3600V to 120V. The transformer is to deliver 1000kW to the factory at 90% efficiency. Find (a) the power delivered to the primary, (1110kW) (b) the current in the primary, and (310A) (c) the current in the secondary. (8300A)

Chapter 22

  1. CH 22 Notes P7 Q5 Light of wavelength 589nm in vacuum passes through a piece of fused quartz of index of refraction n=1.458. (a) Find the speed of light in fused quartz. (2.06×108ms) (b) What is the wavelength of this light in fused quartz? (404nm) (c) What is the frequency of light in the fused quartz? (5.09×1014Hz​)

  2. CH 22 WS #1 Q14 The laws of refraction and reflection are the same for sound as for light. The speed of sound is 340ms in air and 1510ms in water. If a sound wave traveling in air approaches a plane water surface at an angle of incidence of 12.0, what is the angle of refraction? (67.5​)

  3. CH 22 WS #2 Q3 A narrow beam of ultrasonic waves reflects off the liver tumor in the figure. If the speed of the wave is 10.0% less in the liver than in the surrounding medium, determine the depth of the tumor. (6.3cm​) CH 22 WS #2 Q3

  4. CH 22 WS #2 Q6 A submarine is 300m horizontally out from the shore and 100m beneath the surface of the water. A laser beam is sent from the sub so that it strikes the surface of the water at a point 210m from the shore. If the beam just strikes the top of a building standing directly at the water’s edge, find the height of the building. (108m)

Chapter 23

  1. CH 23 Notes P 5 Q3 Assume that a certain concave spherical mirror has a focal length of 10.0cm. (a) Locate the image and find the magnification for an object distance of 25.0cm. Determine whether the image is real or virtual, inverted or upright, and larger or smaller. Do the same for object distances of (16.4cm, 0.667) (b) 10.0cm and (, ) (c) 5.00cm. (10.0cm, 2.00​)

  2. CH 23 WS #1 Q4 A dentist uses a mirror to examine a tooth. The tooth is 1.00cm in front of the mirror, and the image formed 10.0cm behind the mirror. Determine (a) the mirror’s radius of curvature and (2.22cm) (b) the magnification of the image. (10.0​)

  3. CH 23 WS #1 Q6 A dedicated sports car enthusiast polishes the inside and outside surfaces of a hubcap that is a section of a sphere. When he looks into one side of the hubcap, he sees an image of his face 30.0cm in back of the hubcap. He then turns the hubcap over and sees another image of his face 10.0cm in back of the hubcap. (a) How far is his face from the hubcap? (15cm) (b) What is the radius of curvature of the hubcap? (60.0cm​)

  4. CH 23 WS #2 Q7 A man standing 1.52m in front of a shaving mirror produces an inverted image 18.0cm in front of it. How close to the mirror should he stand if he wants to form an upright image of his chin that is twice the chin’s actual size? (8.05cm​)

  5. CH 23 WS #4 Q11 A person looks at a gem with a converging lens with a focal length of 12.5cm. The lens forms a virtual image 30.0cm from the lens. Determine the magnification. Is the image upright or inverted? (3.40​, upright)

  6. CH 23 WS #4 Q13 An object is placed 20cm to the left of a converging lens of focal length 25cm. A diverging lens of focal length 10cm is 25cm to the right of the converging lens. Find the position and magnification of the final image. (9.3cm to left of 2nd lens, M=0.370)

Chapter 24

  1. CH 24 Notes P8 Q5&6 A pair of glass slides 10.0cm long and with n=1.52 are separated on one end by a hair, forming a triangular wedge of air as illustrated in the figure. When coherent light from a helium-neon laser with wavelength 633nm is incident on the film from above, 15.0 dark fringes per centimeter are observed. (a) How thick is the hair? (4.75×105m) (b) The air wedge is replaced with water, with n=1.33. Find the distance between dark bands when the helium-neon laser hits the glass slides. (5.01×104m​)

  2. CH 24 WS #3 Q7 A screen is placed 50.0cm from a single slit, which is illuminated with light of wavelength 680nm. If the distance between the first and third minima in the diffraction pattern is 3.00mm, what is the width of the slit? (0.227mm)

Chapter 26

  1. CH 26 WS #1 Q6 The average lifetime of a pi meson in its own frame of reference is 2.6×108s. (This is the proper lifetime.) If the meson moves with a speed of 0.98c, what is (a) its mean lifetime as measured by an observer on Earth and (1.3×107s) (b) the average distance it travels before decaying as measured by an observer on Earth? (38m) (c) What distance would it travel if time dilation did not occur? (7.6m​)

  2. CH 26 WS #2 Q4 A muon formed high in the Earth’s atmosphere travels at speed v=0.99c for a distance of 4.6km before it decays into an electron, a neutrino, and an anti-neutrino. (a) How long does the muon live, as measured in its reference frame? (2.19×106s) (b) How far does the muon travel, as measure in its frame? (649m​)

  3. CH 26 WS #2 Q5 A spaceship of proper length 300m takes 0.75µs to pass an Earth observer. Determine the speed of this spaceship as measured by the Earth observer. (0.80c)

Chapter 27

  1. CH 27 Notes In converting electrical energy into light energy, a 60W incandescent light bulb operates at about 2.1% efficiency. Assuming that all of the light is green light (vacuum wavelength of green light =555nm), determine the number of photons per second given off by the bulb. (3.6×1018s1​)

  2. CH 27 Notes Light of wavelength 95nm shines on a selenium surface, which was a work function of 5.9eV. The ejected electrons have some kinetic energy. Determine the maximum speed with which electrons are ejected(mass of an electron =9.11×1031kg). (1.6×106ms​)

  3. CH 27 Notes An electron (m=9.11×1031kg) and a proton (m=1.67×1027kg) have the same kinetic energy and are moving at speed much less than the speed of light. Determine the ratio of the de Broglie wavelength of the electron to that of the proton. (42.8​)

  4. CH 27 Summary WS Q3 The threshold of dark-adapted (scotopic) vision is 4.0××1011Wm2 at a central wavelength of 500nm. If light with this intensity and wavelength enters the eye when the pupil is open to its maximum diameter of 8.5mm, how many photons per second enter the eye? (5700​)

  5. CH 27 Summary WS Q5 Ultraviolet light is incident normally on the surface of a certain substance. The binding energy of the electrons in this substance is 3.44eV. The incident light has an intensity of 0.055Wm2 . The electrons are photoelectrically emitted with a maximum speed of 4.2×105ms. How many electrons are emitted from a square centimeter of the surface? Assume that 100% of the photons are absorbed. (melectron=9.11×1031kg) (8.73×1012s1​)

  6. CH 27 Summary WS Q11 The spacing between planes of nickel atoms in a nickel crystal is 0.352nm. At what angle does a second-order Bragg reflection occur in nickel for 11.3keV X-rays? (18.2​)

  7. CH 27 Summary WS Q18 After learning about de Broglie’s hypothesis that particles of momentum p have wave characteristics with wavelength λ=hp, an 80kg student has grown concerned about being diffracted when passing though a 75cm wide doorway. Assuming that significant diffraction occurs when the width of the diffraction aperture is less than 10 times the wavelength of the wave being diffracted, (a) determine the maximum speed at which the student can pass through the doorway in order to be significantly diffracted. (1.1×1034ms) (b) With that speed, how long will it take the student to pass through the doorway if it is 15cm thick? (1.4×1033s) (c) Compare your result to the currently accepted age of the Universe, which is 4×1017s​. Should this student worry about being diffracted? (no)

  8. CH 27 Summary WS Q20 A 50.0g ball moves at 30.0ms. If its speed is measured to an accuracy of 0.10%, what is the minimum uncertainty in its position? (3.5×1032m)

Chapter 28

  1. CH 28 Summary WS Q4 The “size” of the atom in Rutherford’s model is about 1.0×1010m. (melectron=9.11×1031kg) (a) Determine the speed of an electron moving about the proton using the attractive electrostatic force between an electron and a proton separated by this distance. (1.59×106ms) (b) Compute the de Broglie wavelength of the electron as it moves about the proton.(0.457nm​) (c) Does this wavelength suggest that wave effects, such as diffraction and interference, must be considered when studying the atom? (yes)

  2. CH 28 Summary WS Q7 (a) If an electron makes a transition form the n=4 Bohr orbit to the n=2 orbit, determine the wavelength of the photon created in the process. (486nm) (b) Assuming that the atom was initially at rest, determine the recoil speed of the hydrogen atom when this photon is emitted. (0.816ms)

References

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1 Imagine a constant force being applied over a given distance. Let the y axis of a graph be force, and let the x axis be distance. The graph would look like a straight line, y=F. Because W=Fx, we can imagine work to be the area under this line. F, the height, times x, the width. Therefore, we can conclude that, for any force, the work done will be equal to the area under the curve with respect to distance: Fdx. Conveniently, evaluating this integral, we get Fdx=Fx, and W=Fx.
2 Non-conservative forces don't create or destroy energy, rather convert energy to and from usable/relavant forms. For example. The non-conservative work done by friction that would be subtracted from Ei isn't just lost, it's converted into other forms, such as heat energy. However, because we don't measure this type of energy or include it in our calculations because it doesn't fall within mechanical energy (the type of energy we're particularly concerned about), for our relevance, we can consider it lost.
3 Technically, ΔsΔt is the average change in velocity, and to get instantaneous velocity you would have to take the limit as Δt approaches 0: limΔt0ΔsΔt=vt. In the derivation of (40), if you take the limit of one side you must take the limit of the left side, therefore: limΔt0ΔθΔt=ω.
4 The outward force referred to here is actually called centrifugal force. Centrifugal force, however, is a pseudo force; it doesn't actually exist. People assume that it exists because when they see something spinning, they also observe it being pushed outward and conclude that there must be an outward pushing force. In reality, though, this effect is only because of the object's inertia (tendency to not move). Every action has an equal and opposite reaction, so by centripetally pulling on an object to make it go in a circle, you will also feel the object pulling back, resisting the centripetal force.
5 Newton created this equation by observing the behavior of celestial bodies. The equation has no mathematic derivation. This is the reason why many others, such as Einstein, doubted and challenged Newton's ideas. For general purposes, though, this equation works, as proven by the fact that we were able to use it to get to the moon.
6 The actual orbits of planets, according to Kepler's first law, are elliptical. However, usually the elliptical orbits barely differ from perfect curcilar orbits. Therefore, for our calculations, we can assume that the orbits are circular. Even if we derived everything using elliptical orbits, the outcomes would pretty much be the same.
7 For example, if there is a wheel spinning left or right, the direction of the vector of ω will be either up or down. This doesn't seem to make intuitive sense. However, even though there is nothing actually moving in that direction, it is just a fabricated generalization to ease calculations.
8 Streamlines are small lines drawn within the diagram of a vessel to represent the direction of the fluid motion. In this course, streamlines are parallel.
9 Have you ever tried to use your thumb on the end of a hose or faucet to control the water pressure. The pressure increases when you create a smaller hole for the water to come out of because the same mass of water must exit the hose as when there was a bigger opening.
10 You might expect the pressure of a fluid to be high wherever the pipe is thinner/the velocity is higher. However, in fact, if the pressure was lower where the fluid was moving slower, then the fluid flow would want to equalize the pressure, and, to fill the low pressure area, it would start flowing faster. A common misconception is that faster moving fluid exert a higher pressure. For example, wouldn't you get knocked over if hit by fast moving water coming out of a fire hose? It is correct that you get knocked over because of the high pressure of the fluid. However, the pressure of the fluid is not high until you slow the water down by getting is its way, and, the now slow moving water exerts a high pressure onto your body, knocking you over. In fact, the pressure of the fast moving water in the air is approximately atmospheric pressure, but when it's stopped, the pressure increases dramatically.
11 Thermal equilibrium is a state that everything tends toward. Given a closed system, initially many objects within the system may be at different temperatures. However, as time passes, all the objects either heat up or cool down and tend toward the same temperature. Once everything is the same temperature, the objects are in thermal equilibrium.
12 Two objects are in thermal contant with each other if heat can be exchanged between them.
13 The Zeroth Law of Thermodynamics is fairly pointless, as it outlines the extremely obvious. In simple terms, it's basically saying if A=C and B=C, then A=B.
14 Notice the similarity between the specific heat of water and the conversion of cal to J. The specific heat of water can also be written as 1kcalkgC. Water also has an unusually high specific heat. This is beneficial for our survival, as a lot of the heat from the sun is absorbed by the ocean while barely raising the water temperature.
15 All this is saying, is that the heat energy required to go from liquid to gas is much greater than the energy required to go from solid to liquid. This should make sense, when you think about the difference in the movement of particles in solids, liquids, and gases.
16 All natural processes are irreversible. Some are almost reversible. If the process occurs extremely slowly such that everything remains virtually in equilibrium, the process can be considered reversible, but it can never actually reach full reversability. In order for a process to be truly reversible, absolutely no energy can be lost, which is not possible to do in the real world.